# Ncert Solutions For Class 10 Maths Ex 6.4

## Ncert Solutions For Class 10 Maths Chapter 6 Ex 6.4

Question 1:

Let ∆ PQR ~ ∆ STU and their areas be, 64cm2 and 121cm2 respectively. If TU = 15.4, then find QR.

Solution:

Given,

The area of a triangle PQR = 64 cm2

The area of a triangle STU = 121 cm2

TU = 15.4 cm

And ∆ PQR ~ ∆ STU

Therefore, area of triangle PQR/ Area of triangle STU = PQ2 / ST2

= PR2 / SU2 = QR2 / TU2 – – – – – – – – – – (1)

[If the two triangles are similar then the ratio of their areas are said to be equal to the square of the ratio of their corresponding sides]

Therefore, 64 / 121 = QR2 / TU2

=> (8/11)2 = (QR/15.4)2

=> 8/11 = QR / 15.4

=> QR = 8 x 15.4 / 11

=> QR = 8 x 1.4

QR = 11.2 cm

Question 2:

Diagonals of a trapezium PQRS with PR ǁ SR intersect each other at the point A. If PQ = 2RS, then find the ratio of the areas of the triangles PAQ and RAS.

Solution:

PQRS is a trapezium having PQ ǁ SR. The diagonals PR and QS intersect each other at a point A.

In ∆ PAQ and ∆ RAS, we have

1=2(Alternateangles)$\angle 1 = \angle 2(Alternate\:\, angles)$ 3=4(Alternateangles)$\angle 3=\angle 4\, (Alternate\, angles)$ 5=6(Verticallyoppositeangle)$\angle 5=\angle 6\, (Vertically\, opposite\, angle)$

Therefore, ∆ PAQ  ~ ∆ RAS [By AAA similarity criterion]

Now, Area of (∆PAQ) / Area of (∆RAS)

= PQ2 / RS2 [If two triangles are similar then the ratio of their areas are equal to the square of the ratio of their corresponding sides]

= (2RS)2 / RS2 [Therefore, PQ=RS]

Therefore, Area of (∆ PAQ)/ Area of (∆RAS)

= 4RS2/ RS = 4/1

Hence, the required ratio of the area of ∆PAQ and ∆RAS = 4:1

Question 3:

From the given figure, PQR and SQR are two triangles on the same base QR. If PS intersects QR at A, show that the area of triangle PQR / Area of triangle SQR = PA. SA

Solution:

Given,

PQR and SQR are the triangles in which have the same base QR. PS intersects QR at A.

To prove: Area of triangle PQR/ Area of triangle SQR = PA/ SA

Construction: Let us draw two perpendiculars PY and SX on line QR.

Proof:

We know that the area of a triangle = ½ x base x Height

Therefore, areaΔPQRareaΔSQR$\frac{area\Delta PQR}{area\Delta SQR}$ = 12QR×PY12QR×SX$\frac{\frac{1}{2}QR\times PY}{\frac{1}{2}QR\times SX}$

In ∆ PYA and ∆ SXA,

PYA=SXA(Eachequalsto90)$\angle PYA=\angle SXA(Each\, equals\, to\, 90^{\circ})$ PAY=SAX(Verticallyoppositeangles)$\angle PAY= \angle SAX(Vertically\, opposite\, angles)$

ΔPYAΔSXA(ByAAsimilaritycriterionPYSX)$∴ \Delta PYA\sim \Delta SXA(By\, AA\, similarity\, criterion\, ∴ \frac{PY}{SX})$ = PASA$\frac{PA}{SA}$

=> area of triangle PQR/ area of triangle SQR

=> PA/SA

Question 4:

If the areas of two triangles are similar and equal then, prove that they are congruent.

Solution:

Given:

∆ MNO and ∆ XYZ are similar and equal in area.

To prove that: ∆ MNO $\cong$ ∆ XYZ

Proof: Since, ∆ MNO ~ ∆ XYZ

Therefore, Area of (∆ MNO) / Area of (∆ XYZ) = NO2/ YZ2

=> NO2 / YZ2 = 1 [Since the area of triangle MNO = area of triangle XYZ]

=> NO2 / YZ2

=> NO / YZ

Similarly, we can prove that

MN = XY and MO = XZ

Thus, ∆ MNO $\cong$ ∆XYZ [ By SSS criterion of congruence]

Question 5:

X , Y and Z are respectively the mid-points of sides PQ, QR and RP of ΔPQR. Find the ratio of the areas of ΔXYZ and ΔPQR.

Solution:

Given:

X, Y and Z are the mid-points of sides PQ, QR and RA respectively from the ΔPQR.

To Find: area(ΔXYZ) and area(ΔPQR)

Solution: In Δ PQR, we have
Z is the midpoint of PQ (Given)
Y is the midpoint of PR (Given)
So, by the mid-point theorem, we have
ZY || RQ and ZY = ½ RQ
⇒ ZY || RQ and ZY || QX [QX = ½ QR]
∴ QXYZ is parallelogram [Opposite sides of parallelogram are equal and parallel]
Similarly, in Δ ZQX and Δ XYZ, we have
ZQ = XY (Opposite sides of parallelogram QXYZ)
ZX = ZX (Common)
QX = ZY (Opposite sides of parallelogram QXYZ)
∴ Δ ZQX ≅ ΔXYZ
Similarly, we can prove that
Δ PZY ≅ Δ XYZ
Δ YXR ≅ Δ XYZ
If triangles are congruent, then they are equal in area.
So, area(Δ FBD) = area(Δ DEF) …(i)
area(Δ AFE) = area(Δ DEF) …(ii)
and, area(Δ EDC) = area(Δ DEF) …(iii)
Now, area(Δ ABC) = area(Δ FBD) + area(Δ DEF) + area(Δ AFE) + area(Δ EDC) …(iv)
area(Δ ABC) = area(Δ DEF) + area(Δ DEF) + area(Δ DEF) + area(Δ DEF)
⇒ area(Δ DEF) = 1/4area(Δ ABC) [From (i), (ii) and (iii)]
⇒ area(Δ DEF)/area(Δ ABC) = 1/4
Hence, area(Δ DEF):area(Δ ABC) = 1:4

Question 6:

Prove that the ratio of the areas of two similar triangles is equal to the square of the ratio of their corresponding medians.

Solution:

Given:

PA and XB are the medians of triangles PQR and XYZ respectively and Δ PQR ~ Δ XYZ

To prove: Area (Δ PQR)/ Area (Δ XYZ) = PA2 / XB2

Proof: Δ PQR ~ Δ XYZ (Given)

Therefore, Area (Δ PQR) / area (Δ XYZ ) = ( PQ2 / XY2) – – – – – – (1)

And, PQ / XY = QR / YZ = RP / ZX – – – – – (2)

PQXY=12QR12YZ=RXZX$\frac{PQ}{XY}=\frac{\frac{1}{2}QR}{\frac{1}{2}YZ}=\frac{RX}{ZX}$

In Δ PQA and Δ XYB, we have

Therefore, Q=Y(SinceΔPQRΔXYZ)$\angle Q= \angle Y (Since\, \Delta PQR\sim \Delta XYZ)$

PQ / XY = QA / YB [Prove in (1)]

Therefore, Δ PQR ~ Δ XYZ [By SAS similarity criterion]

=> PQ/XY = PA/ XB – – – – – – – (3)

Therefore, Δ PQA ~ Δ XYB

The areas of two similar triangles are proportional to the squares of the corresponding sides.

Therefore, Area of triangle PQR / Area of triangle XYZ = PQ2 / XY2 = PA2/ XB2

Question 7:

Prove that the area of an equilateral triangle described on one side of a square is equal to half the area of the equilateral triangle described on one of its diagonals.

Solution:

Given;

PQRS is a square whose one diagonal is PR. Δ PXR and Δ QYR are two equilateral triangles described on the diagonals PR and side QR of the square PQRS.

To prove:

Area of Δ QYR = ½ Area of Δ PXR

Proof:

Δ PXR and Δ QYR are both equilateral triangles (Given)

Therefore, Δ PXQ ~ Δ QYR [AAA similarity criterion]

Therefore, Area of Δ PXR / Area of Δ QYR = PR2 / QR2

(2QRQR)2=2QR2QR2=2[Since,Diagonal=2side=2QR]$(\frac{\sqrt{2}QR}{QR})^{2}=\frac{2QR^{2}}{QR^{2}}=2[Since,\, Diagonal=\sqrt{2}side=\sqrt{2}QR]$

=> Area (Δ PXR) = 2 x area (Δ QYR)

=> area (Δ QYR) = ½ area (ΔPXR)

Question 8:

Tick the correct solutions and explain.

PQR and QST are two equilateral triangles such that S is the midpoint of QR. The ratio of the areas of triangles PQR and QST is:

(i) 2 : 1

(ii) 1 : 2

(iii) 4 : 1

(iv) 1 : 4

Solution:

Δ PQR and Δ QST are two equilateral triangle. S is the midpoint of QR.

Therefore, QS = SR = ½ QR

Let each side of triangle be 2a

As, Δ PQR ~ Δ QST

Therefore, area (Δ PQR ) / area (Δ QST) = PQ2 / QS2 = (2a)2 / (a)2  = 4a2 / a2  = 4/1 = 4:1

Hence, The correct option is (iii)

Question 9:

Sides of two similar triangles are in the ratio 4 : 9. Areas of these triangles are in the ratio

(i) 2 : 3

(ii) 4: 9

(iii) 81 : 16

(iv) 16 : 81

Solution:

PQR and XYZ are two similarity triangles Δ PQR ~ Δ XYZ (Given)

And, PQ / XY = PR / XZ = QR / YZ = 4/9 (Given)

Therefore, Area of ΔPQR / Area of Δ XYZ = PQ2 / XY 2 [the ratio of the areas of these triangles will be equal to the squares of the ratio of the corresponding sides]

Therefore, Area of Δ PQR / Area of Δ XYZ = (4/9)2 = 16/81

=> 16:81

Hence, the correct option is (iv)