RS Aggarwal Solutions for Class 10, maths Chapter 7

Class 10: Maths Chapter 7 solutions. Complete Class 10 Maths Chapter 7 Notes.

Contents

1 RS Aggarwal Solutions for Class 10 Maths Chapter 7–Triangles
2 RS Aggarwal Solutions for Class 10 Maths Chapter 7: Download PDF
3 Chapterwise RS Aggarwal Solutions for Class 10 Maths :
4 About RS Aggarwal Class 10 Book
5 FAQs
6 Read More

Let the trapezium be ABCD with E and F as the mid points of AD and BC, respectively.
Produce AD and BC to meet at P.

In $△ P A B, D C ∥ A B .$
Applying Thales’ theorem, we get:
$\frac{P D}{D A} = \frac{P C}{C B} Now, E and F are the midpoints of A D and B C, respectively . \Rightarrow \frac{P D}{2 D E} = \frac{P C}{2 C F} \Rightarrow \frac{P D}{D E} = \frac{P C}{C F} Applying the converse of Thales ‘ theorem in ∆ P E F, we get that D C ∥ E F . Hence, E F ∥ A B .$

Thus. EF is parallel to both AB and DC.
This completes the proof.

Page No 373:

Question 7:

In the adjoining figure, ABCD is a trapezium in which CD ∥ AB and its diagonals intersect at O. If AO = (2x + 1) cm, OC = (5x – 7) cm, DO = (7x − 5) cm and OB = (7x + 1) cm, find the value of x.

Answer:

In trapezium ABCD, $A B ∥ C D$ and the diagonals AC and BD intersect at O.
Therefore,

$\frac{A O}{O C} = \frac{B O}{O D} \Rightarrow \frac{2 x + 1}{5 x - 7} = \frac{7 x + 1}{7 x - 5} \Rightarrow (5 x - 7) (7 x + 1) = (7 x - 5) (2 x + 1) \Rightarrow 35 x^{2} + 5 x - 49 x - 7 = 14 x^{2} - 10 x + 7 x - 5 \Rightarrow 21 x^{2} - 41 x - 2 = 0 \Rightarrow 21 x^{2} - 42 x + x - 2 = 0 \Rightarrow 21 x (x - 2) + 1 (x - 2) = 0 \Rightarrow (x - 2) (21 x + 1) = 0 \Rightarrow x = 2, - \frac{1}{21} ∵ x \neq - \frac{1}{21} ∴ x = 2$

Page No 373:

Question 8:

In △ABC, M and N are points on AB and AC respectively such that BM = CN. If ∠B = ∠C then show that MN∥BC.

Answer:

In △ABC, ∠B = ∠C
∴AB = AC (Sides opposite to equal angle are equal)
Subtracting BM from both sides, we get
AB − BM = AC − BM
⇒AB − BM = AC − CN (∵BM = CN)
⇒AM = AN
∴∠AMN = ∠ANM (Angles opposite to equal sides are equal)
Now, In △ABC,
∠A + ∠B + ∠C = 180^o               …..(1)
                                                               (Angle Sum Property of triangle)
Again In In △AMN,
∠A +∠AMN + ∠ANM = 180^o    ……(2)
                                                               (Angle Sum Property of triangle)
From (1) and (2), we get
∠B + ∠C = ∠AMN + ∠ANM
⇒2∠B = 2∠AMN
⇒∠B = ∠AMN
Since, ∠B and ∠AMN are corresponding angles.
∴ MN∥BC

Page No 373:

Question 9:

∆ABC and ∆DBC lie on the same side of BC, as shown in the figure. From a point P on BC, PQ ∥ AB and PR ∥ BD are drawn, meeting AC at Q and CD at R, respectively. Prove that QR ∥ AD.

Answer:

Page No 374:

Question 13:

In ∆ABC, the bisector of ∠B meets AC at D. A line PQ ∥ AC meets AB, BC and BD at P, Q and R respectively.
Show that PR × BQ = QR × BP.

Answer:

In triangle BQP, BR bisects angle B.
Applying angle bisector theorem, we get:
$\frac{Q R}{P R} = \frac{B Q}{B P}$
$\Rightarrow B P \times Q R = B Q \times P R$
This completes the proof.

Page No 399:

Question 1:

Find the pair of similar triangles among the given pairs. State the similarity criterion and write the similarity relation in symbolic form.

(i)

(ii)

(iii)

(iv)

Page No 441:

Question 1:

The sides of certain triangles are given below. Determine which of them are right triangles.
(i) 9 cm, 16 cm, 18 cm
(ii) 7 cm, 24 cm, 25 cm
(iii) 1.4 cm, 4.8 cm, 5 cm
(iv) 1.6 cm, 3.8 cm, 4 cm
(v) $(a - 1) cm, 2 \sqrt{a} cm, (a + 1) cm$

Answer:

Page No 441:

Question 6:

Two vertical poles of height 9 m and 14 m stand on a plane ground. If the distance between their bases is 12 m, find the distance between their tops.

Answer:

Let the two poles be DE and AB and the distance between their bases be BE.
We have:
DE = 9 m, AB = 14 m and BE = 12 m
Draw a line parallel to BE from D, meeting AB at C.
Then, DC = 12 m and AC = 5 m
We need to find AD, the distance between their tops.

Applying Pythagoras theorem in right-angled triangle ACD, we have:

$A D^{2} = A C^{2} + D C^{2} A D^{2} = 5^{2} + 12^{2} = 25 + 144 = 169 A D = \sqrt{169} = 13 m$

Hence, the distance between the tops of the two poles is 13 m.

Page No 442:

Question 7:

A guy wire attached to a vertical pole of height 18 m is 24 m long and has a stake attached to the other end. How far from the base of the pole should the stake be driven so that the wire will be taut?

Answer:

Let AB be a guy wire attached to a pole BC of height 18 m. Now, to keep the wire taut let it to be fixed at A.
Now, In right triangle ABC
By using Pythagoras theorem, we have
${AB}^{2} = {BC}^{2} + {CA}^{2} \Rightarrow 24^{2} = 18^{2} + {CA}^{2} \Rightarrow {CA}^{2} = 576 - 324 \Rightarrow {CA}^{2} = 252 \Rightarrow CA = 6 \sqrt{7} m$
Hence, the stake should be driven $6 \sqrt{7}$ m far from the base of the pole.

Page No 442:

Question 8:

In the given figure, O is a point inside a ∆PQR such that ∠POR = 90°, OP = 6 cm and OR = 8 cm. If PQ = 24 cm and QR = 26 cm, prove that ∆PQR is right-angled.

Answer:

Page No 443:

Question 19:

ABC is an isosceles triangle, right-angled at B. similar triangles ACD and ABE are constructed on sides AC and AB. Find the ratio between the areas of ∆ABE and ∆ACD.

Answer:

$We have, ABC as an isosceles triangle, right angled at B . Now, AB = BC$

Applying Pythagoras theorem in right-angled triangle ABC, we get:

${AC}^{2} = {AB}^{2} + {BC}^{2} = 2 {AB}^{2} (∵ AB = AC)$ …(i)

$∵$ $△ ACD ~ △ ABE$

$We know that ratio of areas of 2 similar triangles is equal to squares of the ratio of their corresponding sides .$

$∴ \frac{ar (△ ABE)}{ar (△ ACD)} = \frac{{AB}^{2}}{{AC}^{2}} = \frac{{AB}^{2}}{2 {AB}^{2}} [from (i)] = \frac{1}{2} = 1 : 2$

Page No 443:

Question 20:

An aeroplane leaves an airport and flies due north at a speed of 1000 km/hr. At the same time, another plane leaves the same airport and flies due west at a speed of 1200 km/hr. How far apart will the two planes be after $1 \frac{1}{2}$ hours?

Answer:

Let A be the first aeroplane flied due north at a speed of 1000 km/hr and B be the second aeroplane flied due west at a speed of 1200 km/hr
Distance covered by plane A in $1 \frac{1}{2}$ hours = $1000 \times \frac{3}{2} = 1500 km$
Distance covered by plane B in $1 \frac{1}{2}$ hours = $1200 \times \frac{3}{2} = 1800 km$
Now, In right triangle ABC
By using Pythagoras theorem, we have
${AB}^{2} = {BC}^{2} + {CA}^{2} = {(1800)}^{2} + {(1500)}^{2} = 3240000 + 2250000 = 5490000 ∴ {AB}^{2} = 5490000 \Rightarrow AB = 300 \sqrt{61} m$
Hence, the distance between two planes after $1 \frac{1}{2}$ hours is $300 \sqrt{61}$ m.

Page No 443:

Question 21:

In △ABC, AD is a median and AL ⊥ BC.
Prove that
$(i) {AC}^{2} = {AD}^{2} + BC . DL + {(\frac{BC}{2})}^{2} (ii) {AB}^{2} = {AD}^{2} - BC . DL + {(\frac{BC}{2})}^{2} (iii) {AB}^{2} + {AC}^{2} = 2 {AD}^{2} - \frac{1}{4} {BC}^{2}$

Answer:

(a)
In right triangle ALD
Using Pythagoras theorem, we have
AL² = AD² − DL² …..(1)
Again, In right triangle ACL
Using Pythagoras theorem, we have
${AC}^{2} = {AL}^{2} + {LC}^{2} = {AD}^{2} - {DL}^{2} + {(DL + DC)}^{2} [Using (1)] = {AD}^{2} - {DL}^{2} + {(DL + \frac{BC}{2})}^{2} [∵ AD is a median] = {AD}^{2} - {DL}^{2} + {DL}^{2} + {(\frac{BC}{2})}^{2} + BC . DL ∴ {AC}^{2} = {AD}^{2} + BC . DL + {(\frac{BC}{2})}^{2} . . . . . (2)$
(b)
In right triangle ALD
Using Pythagoras theorem, we have
AL² = AD² − DL² …..(3)
Again, In right triangle ABL
Using Pythagoras theorem, we have
${AB}^{2} = {AL}^{2} + {LB}^{2} = {AD}^{2} - {DL}^{2} + {LB}^{2} [Using (3)] = {AD}^{2} - {DL}^{2} + {(BD - DL)}^{2} = {AD}^{2} - {DL}^{2} + {(\frac{1}{2} BC - DL)}^{2} = {AD}^{2} - {DL}^{2} + {(\frac{BC}{2})}^{2} - BC . DL + {DL}^{2} ∴ {AB}^{2} = {AD}^{2} - BC . DL + {(\frac{BC}{2})}^{2} . . . . . (4)$
(c) Adding (2) and (4), we get

${AC}^{2} + {AB}^{2} = {AD}^{2} + BC . DL + {(\frac{BC}{2})}^{2} + {AD}^{2} - BC . DL + {(\frac{BC}{2})}^{2} = 2 {AD}^{2} + \frac{{BC}^{2}}{4} + \frac{{BC}^{2}}{4} = 2 {AD}^{2} + \frac{1}{2} {BC}^{2}$

Page No 443:

Question 22:

Naman is doing fly-fishing in a stream. The tip of his fishing rod is 1.8 m above the surface of the water and the fly at the end of the string rests on the water 3.6 m away from him and 2.4 m from the point directly under the tip of the road. Assuming that the string(from the top of his road to the fly) is taut, how much string does he have out (see the adjoining figure)? If he pulls in the string at the rate of 5 cm per second, what will be the horizontal distance of the fly from him after seconds.

Answer:

Naman pulls in the string at the rate of 5 cm per second.
Hence, after 12 seconds the length of the string he will pulled is given by
12 × 5 = 60 cm or 0.6 m
Now, In △BMC
By using Pythagoras theorem, we have
BC² = CM² + MB²
= (2.4)² + (1.8)²
= 9
∴ BC = 3 m
Now, BC’ = BC − 0.6
= 3 − 0.6
= 2.4 m
Now, In △BC’M
By using Pythagoras theorem, we have
C’M² = BC’² − MB²
= (2.4)² − (1.8)²
= 2.52
∴ C’M = 1.6 m
The horizontal distance of the fly from him after 12 seconds is given by
C’A = C’M + MA
= 1.6 + 1.2
= 2.8 m

Page No 447:

Question 12:

Two triangles ABC and PQR are such that AB = 3 cm, AC = 6 cm, ∠A = 70⁰ PR = 9 cm, ∠P = 70⁰ and PQ = 4.5 cm. Show that
△ABC ∼ △PQR and state similarity theorem.

Answer:

Now, In △ABC and △PQR
∠A = ∠P = 70⁰ (Given)
$\frac{AB}{PQ} = \frac{AC}{PR} [∵ \frac{3}{4.5} = \frac{6}{9} \Rightarrow \frac{1}{1.5} = \frac{1}{1.5}]$
By SAS similarity criterion, △ABC ∼ △PQR

Page No 447:

Question 13:

If △ABC ∼ △DEF such that 2AB = DE and BC = 6 cm, find EF

Answer:

When two triangles are similar, then the ratios of the lengths of their corresponding sides are equal.
Here, △ABC ∼ △DEF
$∴ \frac{AB}{DE} = \frac{BC}{EF} \Rightarrow \frac{AB}{2 AB} = \frac{6}{EF} \Rightarrow EF = 12 cm$

Page No 447:

Question 14:

In the given figure, DE∥BC such that AD = x cm, DB = (3x + 4) cm, AE = (x + 3) cm and EC = (3x + 19) cm. Find the value of x

Answer:

Page No 447:

Question 18:

In a trapezium ABCD, it is given that AB∥CD and AB = 2 CD. Its diagonals AC and BD intersect at a point O such that ar(△AOB) = 84 cm²
Find ar(△COD)

Answer:

In △AOB and COD
∠ABO = ∠CDO (Alternte angles in AB∥CD)
∠AOB = ∠COD (Vertically opposite angles)
By AA similarity criterion, △AOB ∼ △COD
If two triangles are similar, then the ratio of their areas is equal to the ratio of the squares of their corresponding sides.
$∴ \frac{area (△ AOB)}{area (△ COD)} = {(\frac{AB}{CD})}^{2} \Rightarrow \frac{84}{area (△ COD)} = {(\frac{2 CD}{CD})}^{2} \Rightarrow area (△ COD) = 12 {cm}^{2}$

Page No 447:

Question 19:

The corresponding sides of two similar triangles are in the ratio of 2 : 3. If the area of the smaller triangle is 48 cm² , find the area of larger triangle.

Answer:

Page No 447:

Question 23:

In the given figure, MN∥BC and AM : MB = 1 : 2.
Find $\frac{area (△ AMN)}{area (△ ABC)}$

Answer:

We have
AM : MB = 1 : 2
$\Rightarrow \frac{MB}{AM} = \frac{2}{1}$
Adding 1 to both sides, we get
$\Rightarrow \frac{MB}{AM} + 1 = \frac{2}{1} + 1 \Rightarrow \frac{MB + AM}{AM} = \frac{2 + 1}{1} \Rightarrow \frac{AB}{AM} = \frac{3}{1}$
Now, In △AMN and △ABC
∠AMN = ∠ABC (Corresponding angles in MN∥BC)
∠ANM = ∠ACB (Corresponding angles in MN∥BC)
By AA similarity criterion, △AMN ∼ △ABC
If two triangles are similar, then the ratio of their areas is equal to the ratio of the squares of their corresponding sides.
$∴ \frac{area (△ AMN)}{area (△ ABC)} = {(\frac{AM}{AB})}^{2} = {(\frac{1}{3})}^{2} = \frac{1}{9}$

Page No 447:

Question 24:

In a triangle BMP and CNR it is given that PB = 5 cm, MP = 6 cm, BM = 9 cm and NR = 9 cm. If △BMP ∼ △CNR, then find the perimeter of the △CNR

Answer:

Page No 448:

Question 28:

In the given figure, ∠AMN = ∠MBC = 76⁰ If p, q and r are the lengths of AM, MB and BC respectively, then express the length of MN in terms of p, q and r.

Answer:

In △AMN and △ABC
∠AMN = ∠ABC = 76⁰ (Given)
∠A = ∠A (Common)
By AA similarity criterion, △AMN ∼ △ABC
If two triangles are similar, then the ratio of their corresponding sides are proportional
$∴ \frac{AM}{AB} = \frac{MN}{BC}$
$\Rightarrow \frac{AM}{AM + MB} = \frac{MN}{BC} \Rightarrow \frac{a}{a + b} = \frac{MN}{c} \Rightarrow MN = \frac{a c}{a + b}$

Page No 448:

Question 29:

The length of the diagonals of a rhombus are 40 cm and 42 cm. Find the length of each sides of the rhombus

Answer:

Suppose ABCD is a rhombus.
We know that the digonals of a rhombus perpendicularly bisect each other.
∴ ∠AOB = 90⁰, AO = 20 cm and BO = 21 cm
Now, In right triangle AOB
By using Pythagoras theorem we have
AB² = AO² + OB²
= 20² + 21²
= 400 + 441
= 841
∴AB² = 841
⇒ AB = 29 cm
Since, all the sides of a rhombus are equal.
Hence, AB = BC = CD = DA = 29 cm

Page No 448:

Question 30:

For each of the following statements state whether true(T) or false(F).
(i) Two circles with different radii are similar.
(ii) Any two rectangles are similar.
(iii) If two traingles are similar, the their corresponding angles are equal and their corresponding sides are equal.
(iv) The length of the line segment joining the mid points of any two sides of a trinagle is equal to the half the length of the third side.
(v) In △ABC, AB = 6 cm, ∠A = 45⁰ and AC = 8 cm and in △DEF, DF = 9 cm, ∠D = 45⁰ and DE = 12 cm, then △ABC ∼ △DEF.
(vi) The polygon formed by joining the mid points of the sides of a quadrilateral is a rhombus.
(vii) The ratio of areas of two similar triangles is equal to the ratio of their corresponding angle-bisector segments.
(viii) The ratio of the perimeters of two similar triangles is same as the ratio of their corresponding medians.
(ix) If O is any point inside a rectangle ABCD then OA² + OC² = OB² + OD²
(x) The sum of the squares on the sides of a rhombus is equal to the sum of the squares on its diagonals.

Answer:

(ii) True
Two circles of any radii are similar to each other.

(i) False
Two rectangles are similar if their corresponding sides are proportional.

(iii) Falase
If two traingles are similar, the their corresponding angles are equal and their corresponding sides are proportional.

(iv) True
Suppose ABC is a triangle and M, N are

Construction: DE is expanded to F sich that EF = DE
To Prove = $DE = \frac{1}{2} BC$
Proof: In △ADE and △CEF
AE = EC        (E is the mid point of AC)
DE = EF        (By construction)
AED = CEF      (Vertically Opposite angle)
By SAS criterion, △ADE ≅ △CEF
CF = AD   (CPCT)
⇒ BD = CF
∠ADE = ∠EFC    (CPCT)
Since, ∠ADE and ∠EFC are alternate angle
Hence, AD ∥ CF and BD ∥ CF
When two sides of a quadrilateral are parallel, then it is a parallelogram
∵DF = BC and BD ∥ CF
∴BDFC is a parallelogram
Hence, DF = BC
⇒ DE + EF = BC
$\Rightarrow DE = \frac{1}{2} BC$

(v) False
In △ABC, AB = 6 cm, ∠A = 45^∘ and AC = 8 cm and in △DEF, DF = 9 cm, ∠D = 45^∘ and DE = 12 cm, then △ABC ∼ △DEF.

In △ABC and △DEF
∠A = ∠D = 45^∘
$\frac{AB}{AC} \neq \frac{DE}{DF}$
So △ABC is not similar to △DEF

(vi) False
The polygon formed by joining the mid points of the sides of a quadrilateral is a parallelogram.

(vii) True

Given: △ABC ∼ △DEF
To Prove = $\frac{Ar (△ ABC)}{Ar (△ DEF)} = {(\frac{AP}{DQ})}^{2}$
Proof: In △ABP and △DEQ
∠BAP = ∠EDQ        (As ∠A = ∠D, so their Half is also equal)
∠B = ∠E         (△ABC ∼ △DEF)
By AA criterion, △ABP ∼ △DEQ
$\frac{AB}{DE} = \frac{AP}{DQ}$            …..(1)
Since, △ABC ∼ △DEF
$∴ \frac{Ar (△ ABC)}{Ar (△ DEF)} = {(\frac{AB}{DE})}^{2} \Rightarrow \frac{Ar (△ ABC)}{Ar (△ DEF)} = {(\frac{AP}{DQ})}^{2} [Using (1)]$

(viii) True

Given: △ABC ∼ △DEF
To Prove = $\frac{Perimeter (△ ABC)}{Perimeter (△ DEF)} = \frac{AP}{DQ}$
Proof: In △ABP and △DEQ
∠B = ∠E          (∵△ABC ∼ △DEF)
∵△ABC ∼ △DEF
$∴ \frac{AB}{DE} = \frac{BC}{EF} \Rightarrow \frac{AB}{DE} = \frac{2 BP}{2 EQ} \Rightarrow \frac{AB}{DE} = \frac{BP}{EQ}$

By SAS criterion, △ABP ∼ △DEQ
$\frac{AB}{DE} = \frac{AP}{DQ}$            …..(1)
Since, △ABC ∼ △DEF
$∴ \frac{Perimeter (△ ABC)}{Perimeter (△ DEF)} = \frac{AB}{DE} \Rightarrow \frac{Perimeter (△ ABC)}{Perimeter (△ DEF)} = \frac{AP}{DQ} [Using (1)]$

(ix) True

Suppose ABCD is a rectangle with O is any point inside it.
Construction: Join OA, OB, OC, OD and draw two parallel lines SQ ∥ AB ∥ DC and PR ∥ BC ∥ AD
To prove: OA² + OC² = OB² + OD²
Proof:
OA² + OC² = (AS² + OS²) + (OQ² + QC²)             [Using Pythagoras theorem in right triangle AOP and COQ]
= (BQ² + OS²) + (OQ² + DS²)
= (BQ² + OQ²) + (OS² + DS²)                                 [Using Pythagoras theorem in right triangle BOQ and DOS]
= OB² + OD²
Hence, LHS = RHS

(x) True

Suppose ABCD is a rhombus having AC and BD its diagonals.
Since, the diagonals of a rhombus perpendicular bisect each other.
Hence, AOC is a right angle triangle
In right triangle AOC
By using Pythagoras theorem, we have
${AB}^{2} = {AO}^{2} + {OB}^{2} \Rightarrow {AB}^{2} = {(\frac{AC}{2})}^{2} + {(\frac{BD}{2})}^{2} [∵ Diagonals of a rhombus perpendicularly bisect each other] \Rightarrow {AB}^{2} = \frac{{AC}^{2}}{4} + \frac{{BD}^{2}}{4} \Rightarrow 4 {AB}^{2} = {AC}^{2} + {BD}^{2} \Rightarrow {AB}^{2} + {AB}^{2} + {AB}^{2} + {AB}^{2} = {AC}^{2} + {BD}^{2} \Rightarrow {AB}^{2} + {BC}^{2} + {CD}^{2} + {DA}^{2} = {AC}^{2} + {BD}^{2} [∵ All sides of a rhombus are equal]$

Page No 451:

Question 7:

In the given figure, O is the point inside a △MNP such that ∠MOP = 90⁰ , OM = 16 cm and OP = 12 cm. If MN = 21 cm and ∠NMP = 90⁰ then
NP = ?
(a) 25 cm
(b) 29 cm
(c) 33 cm
(d) 35 cm

Answer:

Now, In right triangle MOP
By using Pythagoras theorem, we have
${MP}^{2} = {PO}^{2} + {OM}^{2} = 12^{2} + 16^{2} = 144 + 256 = 400 ∴ {MP}^{2} = 400 \Rightarrow MP = 20 cm$
Now, In right triangle MPN
By using Pythagoras theorem, we have
${PN}^{2} = {NM}^{2} + {MP}^{2} = 21^{2} + 20^{2} = 441 + 400 = 841 ∴ {MP}^{2} = 841 \Rightarrow MP = 29 cm$
Hence, the correct answer is option (b).

Page No 451:

Question 8:

The hypotenuse of a right triangle is 25 cm. The other two sides are such that one is 5 cm longer. The lengths of these sides are
(a) 10 cm, 15 cm
(b) 15 cm, 20 cm
(c) 12 cm, 17 cm
(d)13 cm, 18 cm

Answer:

(b) 15 cm, 20 cm

It is given that the length of hypotenuse is 25 cm.
Let the other two sides be x cm and (x $-$ 5) cm.
Applying Pythagoras theorem, we get:

$25^{2} = x^{2} + (x - 5)^{2} \Rightarrow 625 = x^{2} + x^{2} + 25 - 10 x \Rightarrow 2 x^{2} - 10 x - 600 = 0 \Rightarrow x^{2} - 5 x - 300 = 0 \Rightarrow x^{2} - 20 x + 15 x - 300 = 0 \Rightarrow x (x - 20) + 15 (x - 20) = 0 \Rightarrow (x - 20) (x + 15) = 0 \Rightarrow x - 20 = 0 o r x + 15 = 0 \Rightarrow x = 20 or x = - 15 Side of a triangle cannot be negative . Therefore, x = 20 cm$

Now,
x $-$ 5 = 20 $-$ 5 = 15 cm

Page No 451:

Question 9:

The height of an equilateral triangle having each side 12 cm, is
(a) $6 \sqrt{2} cm$
(b) $6 \sqrt{3} cm$
(c) $3 \sqrt{6} cm$
(d) $6 \sqrt{6} cm$

Answer:

(b) $6 \sqrt{3} cm$

Let ABC be the equilateral triangle with AD as its altitude from A.

In right-angled triangle ABD, we have:

$A B^{2} = A D^{2} + B D^{2} A D^{2} = A B^{2} - B D^{2} = 12^{2} - 6^{2} = 144 - 36 = 108 A D = \sqrt{108} = 6 \sqrt{3} cm$

Page No 452:

Question 10:

∆ABC is an isosceles triangle with AB = AC = 13 cm and the length of altitude from A on BC is 5 cm. Then, BC = ?
(a) 12 cm
(b) 16 cm
(c) 18 cm
(d) 24 cm

Answer:

(d) 24 cm

In triangle ABC, let the altitude from A on BC meets BC at D.
We have:
AD = 5 cm, AB = 13 cm and D is the midpoint of BC.
Applying Pythagoras theorem in right-angled triangle ABD, we get:
$A B^{2} = A D^{2} + B D^{2} \Rightarrow B D^{2} = A B^{2} - A D^{2} \Rightarrow B D^{2} = 13^{2} - 5^{2} \Rightarrow B D^{2} = 169 - 25 \Rightarrow B D^{2} = 144 \Rightarrow B D = \sqrt{144} = 12 cm$

Therefore, BC = 2BD = 24 cm

Page No 452:

Question 14:

In a triangle, the perpendicular from the vertex to the base bisects the base. The triangle is
Figure

(a) right-angle
(b) isosceles
(c) scalene
(d) obtuse-angled

Answer:

(b) isosceles

In an isosceles triangle, the perpendicular from the vertex to the base bisects the base.

Page No 452:

Question 15:

In an equilateral triangle ABC, if AD ⊥ BC, then which of the following is true?

(a) 2AB² = 3AD²
(b) 4AB² = 3AD²
(c) 3AB² = 4AD²
(d) 3AB² = 2AD²

Answer:

Page No 453:

Question 19:

In the given figure, ABCD is a trapezium whose diagonals AC and BD intersect at O such that OA = (3x −1) cm, OB = (2x + 1) cm, OC = (5x − 3) cm and OD = (6x − 5) cm. Then, x = ?

(a) 2
(b) 3
(c) 2.5
(d) 4

Answer:

(a) 2

We know that the diagonals of a trapezium are proportional.
Therefore,

$\frac{O A}{O C} = \frac{O B}{O D} \Rightarrow \frac{3 X - 1}{5 X - 3} = \frac{2 X + 1}{6 X - 5} \Rightarrow (3 X - 1) (6 X - 5) = (2 X + 1) (5 X - 3) \Rightarrow 18 X^{2} - 15 X - 6 X + 5 = 10 X^{2} - 6 X + 5 X - 3 \Rightarrow 18 X^{2} - 21 X + 5 = 10 X^{2} - X - 3 \Rightarrow 18 X^{2} - 21 X + 5 - 10 X^{2} + X + 3 = 0 \Rightarrow 8 X^{2} - 20 X + 8 = 0 \Rightarrow 4 (2 X^{2} - 5 X + 2) = 0 \Rightarrow 2 X^{2} - 5 X + 2 = 0 \Rightarrow 2 X^{2} - 4 X - X + 2 = 0 \Rightarrow 2 X (X - 2) - 1 (X - 2) = 0 \Rightarrow (X - 2) (2 X - 1) = 0 \Rightarrow Either x - 2 = 0 or 2 x - 1 = 0 \Rightarrow Either x = 2 or x = \frac{1}{2} When x = \frac{1}{2}, 6 x - 5 = - 2 < 0, which is not possible . Therefore, x = 2$

Page No 453:

Question 20:

The line segments joining the midpoints of the adjacent sides of a quadrilateral form
(a) a parallelogram
(b) a rectangle
(c) a square
(d) a rhombus

Answer:

(a) a parallelogram

The line segments joining the midpoints of the adjacent sides of a quadrilateral form a parallelogram.

Page No 453:

Question 21:

If the bisector of an angle of a triangle bisects the opposite side, then the triangle is
(a) scalene
(b) equilateral
(c) isosceles
(d) right-angled

Answer:

(c) isosceles

Let AD be the angle bisector of angle A in triangle ABC.
Applying angle bisector theorem, we get:
$\frac{A B}{A C} = \frac{B D}{D C}$

It is given that AD bisects BC.
Therefore, BD = DC
$\Rightarrow \frac{A B}{A C} = 1 \Rightarrow A B = A C$

Therefore, the triangle is isosceles.

Answer:

(b) 4:1
In ∆ABC, D is the midpoint of AB and E is the midpoint of AC.
Therefore, by midpoint theorem, DE $∥$ BC.
Also, by Basic Proportionality Theorem,
$\frac{A D}{D B} = \frac{A E}{E C}$

$A l s o, A B = A C = B C (∵ ∆ A B C is an equilateral triangle) S o, \frac{A D}{D B} = \frac{A E}{E C} = 1 In △ A B C a n d △ A D E, we have : ∠ A = ∠ A \frac{A D}{A B} = \frac{A E}{A C} = \frac{1}{2} ∴ △ A B C ~ △ A D E (S A S criterion) ∴ a r (△ A B C) : a r (△ A D E) = {(A B)}^{2} : {(A D)}^{2} \Rightarrow a r (△ A B C) : a r (△ A D E) = 2^{2} : 1^{2} \Rightarrow a r (△ A B C) : a r (△ A D E) = 4 : 1$

Page No 456:

Question 43:

In ∆ABC and ∆DEF, we have: $\frac{A B}{D E} = \frac{B C}{E F} = \frac{A C}{D F} = \frac{5}{7},$ then
ar(∆ABC) : ar(∆DEF) = ?

(a) 5 : 7
(b) 25 : 49
(c) 49 : 25
(d) 125 : 343

Answer:

(b) 25 : 49

$In △ ABC and △ DEF, we have : \frac{A B}{D E} = \frac{B C}{E F} = \frac{A C}{D F} = \frac{5}{7} Therefore, by SSS criterion, we conclude that △ A B C ~ △ D E F . \Rightarrow \frac{a r (△ A B C)}{a r (△ D E F)} = \frac{A B^{2}}{D E^{2}} = {(\frac{5}{7})}^{2} = \frac{25}{49} = 25 : 49$

Page No 456:

Question 47:

If $∆ A B C ~ ∆ Q R P, \frac{ar (∆ A B C)}{ar (∆ P Q R)} = \frac{9}{4},$ AB = 18 cm and BC = 15 cm, then PR = ?
(a) 8 cm
(b) 10 cm
(c) 12 cm
(d) $\frac{20}{3} cm$

Answer:

(b) 10 cm

$∵ △ A B C ~ △ Q R P ∴ \frac{A B}{Q R} = \frac{B C}{P R} Now, \frac{a r (△ A B C)}{a r (△ Q R P)} = \frac{9}{4} \Rightarrow (\frac{A B}{Q R})^{2} = \frac{9}{4} \Rightarrow \frac{A B}{Q R} = \frac{3}{2} Therefore, \frac{A B}{Q R} = \frac{B C}{P R} = \frac{3}{2} Hence, 3 P R = 2 B C = 2 \times 15 = 30 P R = 10 cm$

Page No 456:

Question 48:

In the given figure, O is the point of intersection of two chords AB and CD such that OB = OD and ∠AOC = 45°. Then, ∆OAC and ∆ODB are

(a) equilateral and similar
(b) equilateral but not similar
(c) isosceles and similar
(d) isosceles but not similar

Column I	Column II
(a) In a given ∆ABC, DE ∥ BC and $\frac{A D}{D B} = \frac{3}{5} .$ If AC = 5.6 cm, then AE = …… cm.	(p) 6
(b) If ∆ABC ∼ ∆DEF such that 2AB = 3DE and BC = 6 cm, then EF = …… cm.	(q) 4
(c) If ∆ABC ∼ ∆PQR such that ar(∆ABC) : ar(∆PQR) = 9 : 16 and BC = 4.5 cm, then QR = …… cm.	(r) 3
(d) In the given figure, AB ∥ CD and OA = (2x + 4) cm, OB = (9x − 21) cm, OC = (2x − 1) cm and OD = 3 cm. Then x = ?	(s) 2.1

Answer:

(a) – (s)
Let AE be x.
Therefore, EC = 5.6 $-$ x
It is given that DE $∥$ BC.
Therefore, by B.P.T., we get:

$\frac{A D}{D B} = \frac{A E}{E C} \Rightarrow \frac{3}{5} = \frac{x}{5.6 - x} \Rightarrow 3 (5.6 - x) = 5 x \Rightarrow 16.8 - 3 x = 5 x \Rightarrow 8 x = 16.8 \Rightarrow x = 2.1 cm$

(b) – (q)

$∵ △ A B C ~ △ D E F ∴ \frac{A B}{D E} = \frac{B C}{E F} \Rightarrow \frac{3}{2} = \frac{6}{E F} E F = \frac{6 \times 2}{3} = 4 cm$

(c) – (p)

$∵ △ A B C ~ △ P Q R ∴ \frac{a r (△ A B C)}{a r (△ P Q R)} = \frac{B C^{2}}{Q R^{2}} \Rightarrow \frac{9}{16} = \frac{4 . 5^{2}}{Q R^{2}} \Rightarrow Q R = \sqrt{\frac{4.5 \times 4.5 \times 16}{9}} = \frac{4.5 \times 4}{3} = 6 cm$

(d) – (r)

$∵ A B ∥ C D ∴ \frac{O A}{O B} = \frac{O C}{O D} (Thales ‘ theorem) \Rightarrow \frac{2 x + 4}{9 x - 21} = \frac{2 x - 1}{3} \Rightarrow 3 (2 x + 4) = (2 x - 1) (9 x - 21) \Rightarrow 6 x + 12 = 18 x^{2} - 42 x - 9 x + 21 \Rightarrow 18 x^{2} - 57 x + 9 = 0 \Rightarrow 6 x^{2} - 19 x + 3 = 0 \Rightarrow 6 x^{2} - 18 x - x + 3 = 0 \Rightarrow (6 x - 1) (x - 3) = 0 \Rightarrow x = 3 o r x = - \frac{1}{6} But x = - \frac{1}{6} makes (2 x - 1) < 0, which is not possible . Therefore, x = 3$

Page No 458:

Question 54:

Match the following columns:

Column I	Column II
(a) A man goes 10 m due east and then 20 m due north. His distance from the starting point is …… m.	(p) $25 \sqrt{3}$
(b) In an equilateral triangle with each side 10 cm, the altitude is …… cm.	(q) $5 \sqrt{3}$
(c) The area of an equilateral triangle having each side 10 cm is …… cm².	(r) $10 \sqrt{5}$
(d) The length of diagonal of a rectangle having length 8 m and breadth 6 m is …… m .	(s) 10

Answer:

(a) – (r)
Let the man starts from A and goes 10 m due east at B and then 20 m due north at C.
Then, in right-angled triangle ABC, we have:
$A B^{2} + B C^{2} = A C^{2} \Rightarrow A C = \sqrt{10^{2} + 20^{2}} = \sqrt{100 + 200} = 10 \sqrt{3}$
Hence, the man is 10 $\sqrt{3}$ m away from the starting point.

(b) – (q)
Let the triangle be ABC with altitude AD.
In right-angled triangle ABD, we have:
$A B^{2} = A D^{2} + B D^{2} \Rightarrow A D^{2} = 10^{2} - 5^{2} (∵ B D = \frac{1}{2} B C) \Rightarrow A D = \sqrt{100 - 25} = \sqrt{75} = 5 \sqrt{3} cm$

(c) – (p)
Area of an equilateral triangle with side a = $\frac{\sqrt{3}}{4} a^{2} = \frac{\sqrt{3}}{4} \times 10^{2} = \sqrt{3} \times 5 \times 5 = 25 \sqrt{3} {cm}^{2}$

(d) – (s)
Let the rectangle be ABCD with diagonals AC and BD.
In right-angled triangle ABC, we have:
$A C^{2} = A B^{2} + B C^{2} = 8^{2} + 6^{2} = 64 + 36 \Rightarrow A C = \sqrt{100} = 10 m$

Page No 462:

Question 4:

The areas of two similar triangles are 25 cm² and 36 cm² respectively. If the altitude of the first triangle is 3.5 cm, then the corresponding altitude of the other triangle is
(a) 5.6 cm
(b) 6.3 cm
(c) 4.2 cm
(d) 7 cm

Answer:

(c)
We know that the ratio of areas of similar triangles is equal to the ratio of squares of their corresponding altitudes.
Let h be the altitude of the other triangle.
Therefore,

$\frac{25}{36} = \frac{{(3.5)}^{2}}{h^{2}}$ ^{$\Rightarrow h^{2} = \frac{{(3.5)}^{2} \times 36}{25} \Rightarrow h^{2} = 17.64 \Rightarrow h = 4.2 cm$}

Page No 463:

Question 5:

If ∆ABC ∼ ∆DEF such that 2 AB = DE and BC = 6 cm, find EF.

Answer:

$∵$ ∆ABC ∼ ∆DEF

$∴ \frac{A B}{D E} = \frac{B C}{E F} \Rightarrow \frac{1}{2} = \frac{6}{E F} \Rightarrow E F = 12 cm$

Page No 463:

Question 6:

In the given figure, DE ∥ BC such that AD = x cm, DB = (3x + 4) cm, AE = (x + 3) cm and EC = (3x + 19) cm. Find the value of x.

Answer:

Page No 463:

Question 10:

In an trapezium ABCD, it is given that AB ∥ CD and AB = 2CD. Its diagonals AC and BD intersect at the point O such that ar(∆AOB) = 84 cm². Find ar(∆COD).

Answer:

In ∆AOB and ∆COD, we have:

$∠ A O B = ∠ C O D (Vertically opposite angles) ∠ OAB = ∠ OCD (Alternate angles as AB ∥ CD) Applying AA similiarity criterion, we get : △ AOB ~ △ COD ∴ \frac{a r (△ A O B)}{a r (△ C O D)} = \frac{A B^{2}}{C D^{2}} \Rightarrow \frac{84}{a r (△ C O D)} = {(\frac{A B}{C D})}^{2} \Rightarrow \frac{84}{a r (△ C O D)} = {(\frac{2 C D}{C D})}^{2} \Rightarrow a r (△ C O D) = \frac{84}{4} = 21 {cm}^{2}$

Page No 463:

Question 11:

The corresponding sides of two similar triangles are in the ratio 2 : 3. If the area of the smaller triangle is 48 cm², find the area of the larger triangle.

Answer:

It is given that the triangles are similar.
Therefore, the ratio of areas of similar triangles will be equal to the ratio of squares of their corresponding sides.

$∴ \frac{48}{Area of larger triangle} = \frac{2^{2}}{3^{2}} \Rightarrow \frac{48}{Area of larger triangle} = \frac{4}{9} \Rightarrow Area of larger triangle = \frac{48 \times 9}{4} = 108 {cm}^{2}$

Page No 463:

Question 12:

In the given figure, LM CB and LN CD.
Prove that $\frac{A M}{A B} = \frac{A N}{A D} .$

Answer:

Page No 464:

Question 17:

In the given figure, ∆ABC and ∆DBC have the same base BC. If AD and BC intersect at O, prove that $\frac{ar (∆ A B C)}{ar (∆ D B C)} = \frac{A O}{D O} .$

Answer:

$Construction : Draw AX ⊥ CO and DY ⊥ BO . As, \frac{ar (△ ABC)}{ar (△ DBC)} = \frac{\frac{1}{2} \times AX \times BC}{\frac{1}{2} \times DY \times BC} \Rightarrow \frac{ar (△ ABC)}{ar (△ DBC)} = \frac{AX}{DY} . . . (i) In △ ABC and △ DBC, ∠ AXY = ∠ DYO = 90 ° (By construction) ∠ AOX = ∠ DOY (Vertically opposite angles) ∴ △ AXO \sim △ DYO (By AA criterion) ∴ \frac{AX}{DY} = \frac{AO}{DO} (Thales ‘ s theorem) . . . (ii) From (i) and (ii), we have : \frac{ar (△ ABC)}{ar (△ DBC)} = \frac{AX}{DY} = \frac{AO}{DO} or, \frac{ar (△ ABC)}{ar (△ DBC)} = \frac{AO}{DO}$
This completes the proof.

Page No 464:

Question 18:

In the given figure, XY ∥ AC and XY divides ∆ABC into two regions, equal in area. Show that $\frac{A X}{A B} = \frac{(2 - \sqrt{2})}{2} .$

Answer:

$In △ A B C and △ B X Y, we have : ∠ B = ∠ B ∠ B X Y = ∠ B A C (Corresponding angles) Thus, △ A B C ~ △ B X Y (A A c r i t e r i o n) ∴ \frac{a r (△ A B C)}{a r (△ B X Y)} = \frac{A B^{2}}{B X^{2}} = \frac{A B^{2}}{{(A B - A X)}^{2}} . . . (i) Also, \frac{a r (△ A B C)}{a r (△ B X Y)} = \frac{2}{1} {∵ a r (△ B X Y) = a r (t r a p e z i u m A X Y C)} . . . (ii) From (i) and (ii), we have : \frac{A B^{2}}{{(A B - A X)}^{2}} = \frac{2}{1} \Rightarrow \frac{A B}{(A B - A X)} = \sqrt{2} \Rightarrow \frac{(A B - A X)}{A B} = \frac{1}{\sqrt{2}} \Rightarrow 1 - \frac{A X}{A B} = \frac{1}{\sqrt{2}} \Rightarrow \frac{A X}{A B} = 1 - \frac{1}{\sqrt{2}} = \frac{\sqrt{2} - 1}{\sqrt{2}} = \frac{(2 - \sqrt{2})}{2}$

Page No 464:

Question 19:

In the given figure, ∆ABC is an obtuse triangle, obtuse-angles at B. If AD ⊥ CB, Prove that AC² = AB² + BC² + 2BC ⋅ BD.

Answer:

Applying Pythagoras theorem in right-angled triangle ADC, we get:

$A C^{2} = A D^{2} + D C^{2} \Rightarrow A C^{2} - #160; D C^{2} = A D^{2} \Rightarrow A D^{2} = A C^{2} - D C^{2} . . . (1)$

Applying Pythagoras theorem in right-angled triangle ADB, we get:

$A B^{2} = A D^{2} + D B^{2} \Rightarrow A B^{2} - D B^{2} = A D^{2} \Rightarrow A D^{2} = A B^{2} - D B^{2} . . . (2)$

$From equation (1) and (2), we have : A C^{2} - D C^{2} = A B^{2} - D B^{2} \Rightarrow A C^{2} = A B^{2} + D C^{2} - D B^{2} \Rightarrow A C^{2} = A B^{2} + (D B + B C)^{2} - D B^{2} (∵ D B + B C = D C) \Rightarrow A C^{2} = A B^{2} + D B^{2} + B C^{2} + 2 D B . B C - D B^{2} \Rightarrow A C^{2} = A B^{2} + B C^{2} + 2 B C . B D$

This completes the proof.

Page No 464:

Question 20:

In the given figure, PA, QB and RC are perpendicular to AC. If AP = x, QB = z, RC = y, AB = a and BC = b, show that $\frac{1}{x} + \frac{1}{y} = \frac{1}{z} .$

Answer:

$In △ P A C and △ Q B C, we have : ∠ A = ∠ B (Both angles are 90 °) ∠ P = ∠ Q (Correspond ing angles) and ∠ C = ∠ C (Common angles) Therefore, △ P A C ~ △ Q B C \frac{A P}{B Q} = \frac{A C}{B C}$

$\Rightarrow \frac{x}{z} = \frac{a + b}{b} \Rightarrow a + b = \frac{b x}{z} . . . (1)$

$In △ R C A and △ Q B A, we have : ∠ C = ∠ B (Both angles are 90 °) ∠ R = ∠ Q (Correspond ing angles) and ∠ A = ∠ A (Common angles) Therefore, △ R C A ~ △ Q B A \frac{R C}{B Q} = \frac{A C}{A B} \Rightarrow \frac{y}{z} = \frac{a + b}{a} \Rightarrow a + b = \frac{a y}{z} . . . (2)$
From equation (1) and (2), we have:

$\frac{b x}{z} = \frac{a y}{z} \Rightarrow b x = a y \Rightarrow \frac{a}{b} = \frac{x}{y} . . . (3) Also, \frac{x}{z} = \frac{a + b}{b} \Rightarrow \frac{x}{z} = \frac{a}{b} + 1 Using the value of \frac{a}{b} from equation (3), we have : \Rightarrow \frac{x}{z} = \frac{x}{y} + 1 Dividing both sides by x, we get : \frac{1}{z} = \frac{1}{y} + \frac{1}{x}$

$∴ \frac{1}{x} + \frac{1}{y} = \frac{1}{z} This completes the proof .$

RS Aggarwal Solutions for Class 10 Maths Chapter 7: Download PDF

RS Aggarwal Solutions for Class 10 Maths Chapter 7–Triangles

Download PDF: RS Aggarwal Solutions for Class 10 Maths Chapter 7–Triangles PDF

Chapterwise RS Aggarwal Solutions for Class 10 Maths :

About RS Aggarwal Class 10 Book

Investing in an R.S. Aggarwal book will never be of waste since you can use the book to prepare for various competitive exams as well. RS Aggarwal is one of the most prominent books with an endless number of problems. R.S. Aggarwal’s book very neatly explains every derivation, formula, and question in a very consolidated manner. It has tonnes of examples, practice questions, and solutions even for the NCERT questions.

He was born on January 2, 1946 in a village of Delhi. He graduated from Kirori Mal College, University of Delhi. After completing his M.Sc. in Mathematics in 1969, he joined N.A.S. College, Meerut, as a lecturer. In 1976, he was awarded a fellowship for 3 years and joined the University of Delhi for his Ph.D. Thereafter, he was promoted as a reader in N.A.S. College, Meerut. In 1999, he joined M.M.H. College, Ghaziabad, as a reader and took voluntary retirement in 2003. He has authored more than 75 titles ranging from Nursery to M. Sc. He has also written books for competitive examinations right from the clerical grade to the I.A.S. level.

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