* **Question 1: *

*Let **∆ PQR ~ **∆ STU and their areas be, 64cm ^{2} and 121cm^{2} respectively. If TU = 15.4, then find QR.*

**Solution: **

Given,

The area of a triangle PQR = 64 cm^{2}

The area of a triangle STU = 121 cm^{2}

TU = 15.4 cm

And ∆ PQR ~ ∆ STU

Therefore, area of triangle PQR/ Area of triangle STU = PQ^{2} / ST^{2}

= PR^{2} / SU^{2} = QR^{2} / TU^{2} – – – – – – – – – – (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 = QR^{2} / TU^{2}

=> (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

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

Now, Area of (∆PAQ) / Area of (∆RAS)

= PQ^{2} / RS^{2 }[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 }/ RS^{2 }[Therefore, PQ=RS]

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

= 4RS^{2}/ 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,

In ∆ PYA and ∆ SXA,

=> 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

Proof: Since, ∆ MNO ~ ∆ XYZ

Therefore, Area of (∆ MNO) / Area of (∆ XYZ) = NO^{2}/ YZ^{2}

=> NO^{2} / YZ^{2} = 1 [Since the area of triangle MNO = area of triangle XYZ]

=> NO^{2 }/ YZ^{2}

=> NO / YZ

Similarly, we can prove that

MN = XY and MO = XZ

Thus, ∆ MNO

*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) = PA^{2} / XB^{2 }

Proof: Δ PQR ~ Δ XYZ (Given)

Therefore, Area (Δ PQR) / area (Δ XYZ ) = ( PQ^{2} / XY^{2}) – – – – – – (1)

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

In Δ PQA and Δ XYB, we have

Therefore,

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 = PQ^{2} / XY^{2} = PA^{2}/ XB^{2}

* *

*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 = PR^{2} / QR^{2 }

=> 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) = PQ^{2} / QS^{2} = (2a)^{2 }/ (a)^{2 } = 4a^{2 }/ a^{2 } = 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 = PQ^{2} / 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)