# Class 9 RD Sharma Solutions – Chapter 15 Areas of Parallelograms and Triangles- Exercise 15.3 | Set 1

### Question 1. In the figure, compute the area of quadrilateral ABCD.

**Solution: **

According to the question

DC = 17 cm, AD = 9 cm and BC = 8 cm

Find: the area of quadrilateral ABCD

In Î”BCD,

Using Pythagoras Theorem

CD

^{2}= BD^{2}+ BC^{2}17

^{2}= BD^{2}+ 8^{2}BD

^{2}= 289 âˆ’ 64BD = 15

So, the area of Î”BCD = 1/2(8 Ã— 17) = 68

In Î”ABD

Using Pythagoras Theorem

AB

^{2}+ AD^{2}= BD^{2}15

^{2}= AB^{2}+ 9^{2}AB

^{2}= 225 âˆ’ 81 = 144AB = 12

So, the area of Î”ABD = 1/2(12 Ã— 9) = 54

Now we find the ar(quad ABCD) = ar(Î”ABD) + ar(Î”BCD)

ar(quad ABCD) = 54 + 68 = 122 cm

^{2}

Hence, the area of quadrilateral ABCD is 122 cm^{2}

### Question 2. In the figure, PQRS is a square and T and U are, respectively, the midpoints of PS and QR. Find the area of Î”OTS if PQ = 8 cm.

**Solution: **

According to the question

T and U are mid-points of PS and QR

Hence, TU âˆ¥ PQ

PQ = 8 cm

Find: the area of Î”OTS

In Î”PQS,

It is given that T is the midpoint of PS and TO âˆ¥ PQ

therefore, TO = (1/2) PQ = 4 cm

and TS = (1/2) PS = 4 cm

So, the area of ar(Î”OTS) = (1/2)(TO Ã— TS)

= (1/2)(4 Ã— 4) = 8 cm

^{2}

Hence, the area of Î”OTS is 8 cm^{2}

### Question 3. Compute the area of trapezium PQRS in figure

**Solution: **

According to the figure

PQ = 16 cm

Here T is the mid-point of side PQ so

PT = QT = 8 cm

SR = 8 cm

RQ = 17 cm

Find: the area of trapezium PQRS

So, the ar(trap. PQRS) = ar(rect. PSRT) + ar(Î”QRT) ….(1)

So, In Î”QRT

Using Pythagoras Theorem

QR

^{2}= QT^{2}+ RT^{2}RT

^{2}= QR^{2}âˆ’ QT^{2}RT

^{2}= 17^{2}âˆ’ 8^{2}= 225RT = 15

So, the area of Î”QRT

= 1/2(QT Ã— RT) = 8 Ã— 15 = 180/2 = 60

Now we find the area of rectangle PSRT

= PT Ã— RT = 8 Ã— 15 = 120

Now put all these values in eq(1), we get

ar(trap. PQRS) = 120 + 60 = 180

Hence, the area of trapezium is 180 cm^{2}

### Question 4. In figure, âˆ AOB = 90Â°, AC = BC, OA = 12 cm and OC = 6.5 cm. Find the area of Î”AOB.

**Solution:**

According to the question

âˆ AOB = 90Â°, AC = BC, OA = 12 cm and OC = 6.5 cm

Find: the area of Î”AOB

From the figure, C is the mid-point of hypotenuse AB so in right triangle

the mid-point is equidistant from the vertices

therefore, CA = CB = OC

CA = CB = 6.5 cm

AB = 13 cm

In Î”OAB,

Using Pythagoras Theorem

AB

^{2}= OB^{2}+ OA^{2}13

^{2}= OB^{2}+ 12^{2}OB

^{2}= 13^{2}âˆ’ 12^{2}= 169 âˆ’ 144 = 25OB = 5

Now we find the area of Î”AOB = (1/2)(12 Ã— 5) = 30 cm

^{2}

Hence, the area of Î”AOB is 30 cm^{2}

### Question 5. In figure, ABCD is a trapezium in which AB = 7 cm, AD = BC = 5 cm, DC = x cm, and distance between AB and DC is 4 cm. Find the value of x and area of trapezium ABCD.

**Solution: **

According to the question

AB = 7 cm, AD = BC = 5 cm

Construction: Draw AL âŠ¥ DC, BM âŠ¥ DC,

AL = BM = 4 cm and LM = 7 cm

Find: the value of x and area of trapezium ABCD.

In Î” ADL,

Using Pythagoras Theorem

AD

^{2}= AL^{2}+ DL^{2}25 = 16 + DL

^{2}DL = 3 cm

Similarly,

MC = âˆšBC

^{2}– BM^{2}= âˆš25 – 16 = 3 cmSo, x = CD = CM + ML + LD = (3 + 7 + 3) cm = 13 cm

Now we find the area of trapezium ABCD = 1/2(AB + CD) Ã— AL

= 1/2(7 + 13) Ã— 4 cm

^{2}= 40 cm^{2}

Hence, the value of x is 3 cm and the area of trapezium is 40 cm^{2}

### Question 6. In figure, OCDE is a rectangle inscribed in a quadrant of a circle of radius 10 cm. If OE = 2âˆš5 cm, find the area of the rectangle.

**Solution: **

According to the question

OD = 10 cm and OE = 2âˆš5cm

Find: the area of the rectangle OCDE

So, In Î”DEO

By using Pythagoras theorem

OD

^{2}= OE^{2}+ DE^{2}DE = âˆšOD

^{2}– OE^{2}= âˆš10

^{2}– (2âˆš5)^{2}= 4âˆš5 cmNow we find the area of rectangle OCDE = OE Ã— DE

= 2âˆš5 x 4âˆš5 cm

^{2}= 40 cm^{2}

Hence, the area of rectangle is 40 cm^{2}

### Question 7. In figure, ABCD is a trapezium in which AB âˆ¥ DC. Prove that ar(Î”AOD) = ar(Î”BOC)

**Solution: **

According to the question

ABCD is a trapezium in which AB âˆ¥ DC

To prove : ar(Î”AOD) = ar(Î”BOC)

Proof :

According to the figure, we conclude that Î”ADC and Î”BDC are on the

same base DC and between same parallels AB and DC

So, ar(Î”ADC) = ar(Î”BDC) ….(1)

Here, ar(Î”ADC) = ar(Î”AOD) + ar(Î”DOC)

Similarly, ar(Î”BDC) = ar(Î”BOC) + ar(Î”DOC)

Now put all these values in eq(1), we get

ar(Î”AOD) + ar(Î”DOC) = ar(Î”BOC) + ar(Î”DOC)

ar(Î”AOD) = ar(Î”BOC)

Hence, proved

### Question 8. In figure, ABCD, ABFE and CDEF are parallelograms. Prove that ar(Î”ADE) = ar(Î”BCF).

**Solution: **

According to the question

ABCD is parallelogram, so, AD = BC

CDEF is parallelogram, so, DE = CF

ABFE is parallelogram, so, AE = BF

Prove: ar(Î”ADE) = ar(Î”BCF)

Proof:

In Î”ADF and BCF,

AD = BC,

DE = CF

AE = BF

So, by SSS congruence,

Î”ADE â‰… Î”BCF

So, ar(Î”ADE) = ar(Î”BCF)

Hence, proved

### Question 9. Diagonals AC and BD of a quadrilateral ABCD intersect each other at P. Show that: ar(Î”APB) Ã— ar(Î”CPD) = ar(Î”APD) Ã— ar(Î”BPC).

**Solution: **

Construction : Draw BQ âŠ¥ AC and DR âŠ¥ AC

Prove: ar(Î”APB) Ã— ar(Î”CPD) = ar(Î”APD) Ã— ar(Î”BPC)

Proof:

Lets take L.H.S

= ar(Î”APB) Ã— ar(Î”CDP)

= (1/2) [(AP Ã— BQ)] Ã— (1/2 Ã— PC Ã— DR)

= (1/2 Ã— PC Ã— BQ) Ã— (1/2 Ã— AP Ã— DR)

= ar(Î”APD) Ã— ar(Î”BPC)

L.H.S = R.H.S

Hence proved

### Question 10. In figure, ABC and ABD are two triangles on the base AB. If line segment CD is bisected by AB at O, show that ar(Î”ABC) = ar(Î”ABD).

**Solution: **

According to the question

ABC and ABD are two triangles on the base AB

and CD is bisected by AB at O,

To prove: ar(Î”ABC) = ar(Î”ABD).

Construction: CP âŠ¥ AB and DQ âŠ¥ AB.

Proof:

First we find the area of Î”ABC = 1/2 Ã— AB Ã— CP …..(i)

Now we find the area of Î”ABD = 1/2 Ã— AB Ã— DQ …..(i)

So, in Î”CPO and Î”DQO,

âˆ CPO = âˆ DQO (Each 90Â°)

CO = OD (Given)

âˆ COP = âˆ DOQ (Vertically opposite angles are equal)

So, by AAS congruency

Î”CPO â‰… Î”DQO

So, by C.P.C.T

CP = DQ …..(iii)

So, from equation (i), (ii), and (iii)

ar(Î”ABC) = ar(Î”ABD)

Hence proved

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