The document Short Answer Type Questions- Areas of Parallelograms and Triangles Class 9 Notes | EduRev is a part of the Class 9 Course Class 9 Mathematics by Full Circle.

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**Question 1. PQRS is a square. T and U are respectively, the mid points of PS and QR. Find the area of **Î”**OTS, if PQ = 8 cm, where O is the point of intersection of TU and OS. Solution**. We have, PS = PQ = 8 cm and TU || PQ

âˆµ

ST = (1/2) PS

âˆ´ ST = (1/2)Ã— 8 = 4 cm

Also, PQ = TU = 8 cm

âˆ´ OT = (1/2)TU = (1/2) Ã— 8 = 4 cm

Now, Area of Î”OTS

= (1/2)Ã— ST Ã— OT [âˆµ OTS is a rt. triangle]

= (1/2)Ã— 4 Ã— 4 cm^{2}

= 8 cm^{2}

**Question 2. The area of the parallelogram PQRS (in the adjoining figure) is 88 cm ^{2}. Find the value of x. Solution: **PQRS is a parallelogram.

Area of a parallelogram = base x height

âˆ´ *x* * 8 cm = 88 cm^{2}

â‡’ x= (88/8) cm

â‡’ x= 11 cm Thus, the required value of x is 11 cm.

**Question 3. Find the area of Î”ABC given in the adjoining figure. Solution:** âˆµThe area of a triangle = (1/2)x base x altitude

âˆ´ Area of Î”ABC = (1/2)x 10 cm x 4 cm

= 20 cm^{2}

**Question 4. In the adjoining figure, ABCD is a parallelogram and BPC is a triangle. If the area of parallelogram ABCD = 26 cm ^{2}, then find the area of triangle BPC. Solution:** Since parallelogram ABCD and Î”BPC are on the same base BC and between the same parallels BC and AD.

âˆ´ Area of Î”BPC = (1/2) x Area of parallelogram ABCD

â‡’ Area of Î”BPC =(1/2) x 26 cm^{2 }

= 13 cm^{2}

âˆ´ The required area of Î”BPC = 13 cm^{2}.

**Question 5. In the adjoining figure, PQRS is a parallelogram and PQT is a triangle. If area of triangle PQT = 18 cm ^{2}, then find the area of the parallelogram PQRS.**

**Solution:** âˆµ Parallelogram PQRS and Î”PQT are on the same base PQ and between the same parallels PQ and SR.

âˆ´ Area of parallelogram PQRS = 2(ar Î”PQT)

â‡’ ar (parallelogram PQRS) = 2(18 cm^{2})

â‡’ ar (parallelogram PQRS)

= 36 cm^{2}

Thus, the required area of parallelogram PQRS is 36 cm^{2}.

**Question 6. In the adjoining figure. ABC is triangle and AD is a median. If the area of Î”ABD is 15 cm ^{2}, then find the area of Î”ABC. Solution: **Since, a median divides the triangle into two triangles of equal areas.

âˆ´ ar Î”ABD = (1/2)(ar Î”ABC)

â‡’ 2(ar Î”ABD) = ar (Î”ABC)

â‡’ 2(15 cm^{2}) = ar (Î”ABC)

â‡’ 30 cm^{2} = ar (Î”ABC)

Thus, the required area of Î”ABC is 30 cm^{2}.

**Question 1. The area of **Î”**ABC, in the adjoining figure, is 32 cm ^{2}. AD is a median and E is the mid-point of AD. Find the area of **Î”

Solution:

âˆ´ ar (Î”ABD) =(1/2)x ar (Î”ABC) ...(1)

Similarly, ar (Î”BED) = (1/2) x ar (Î”ABD) ...(2)

From (1) and (2), we have ar (Î”BED)

= (1/2)x (1/2)x [ar (Î”ABC)]

= (1/4) ar (Î”ABC)

= (1/4)x 32 cm^{2 }= 8 cm^{2}

Thus, the area of Î”BED is 8 cm^{2}.

**Question 2 In the adjoining figure, the area of **Î”**BCE is 21 cm ^{2}. If CD = 6 cm, then find the length of AF. Solution:** âˆµ Parallelogarm ABCD and Î”BCE are on the same base and between the same parallels.

âˆ´ ar (Î”BCE) = (1/2) x ar (parallelogram ABCD)

â‡’ 21 cm^{2} = (1/2)x ar (parallelogram ABCD)

â‡’ 21 cm^{2} = (1/2) [CD x AF]

â‡’ 21 cm^{2} = (1/2)[6 x AF]

â‡’ AF = ((21 x 2)/2) cm = 7 cm

Thus, the required length of AF is 7 cm.

**Question 3 ABCD is a parallelogram and BC is produced to a point Q such that AD = CQ. If AQ intersects DC at P, show that: ar (BPC) = ar (DPQ) Solution.** âˆµ The triangles on the same base and between same parallels.

âˆ´ ar (Î”ACP) = ar (Î”BPC) .. (1)

and ar (Î”ADQ) = ar (ADC) ... (2)

â‡’ ar (Î”ADC) - ar (Î”ADP)

= ar (Î”ADQ) - ar (Î”ADP)

â‡’ ar (Î”APC) = ar (Î”APQ) ... (3)

From (1) and (3), we get

ar (Î”BCP) = ar (Î”DPQ)

**Question 4. In the adjoining figure, the area of a parallelogram ABCD is 40 cm ^{2}. If PQ is a median of **Î”

Solution:

= (1/2) (40 cm^{2}) = 20 cm^{2 }

âˆµ PQ is a median of Î”CDP.

âˆ´ ar (Î”PDQ) = (1/2) x ar (Î”CDP)

â‡’ ar (Î”PDQ) = (1/2) x (20 cm^{2}) = 10 cm^{2 }

Thus, the required area of Î”PDQ is 10 cm^{2}.

**Question 5. In the adjoining figure, ABC is a triangle having area as 24 cm ^{2}. Find the area of (i) **Î”

Solution:

âˆ´ ar (Î”DEF) = (1/4) (ar Î”ABC)

ar (Î”DEF) = (1/4)(24 cm^{2}) = 6 cm^{2}

Thus, the required area of Î”DEF = 6 cm^{2}.

ar (parallelogram BDEF) = ar (Î”BDF) + ar (Î”DEF)

= (1/4)x ar (Î”ABC) + (1/4)x ar (Î”ABC)

= (1/4)x (24 cm^{2}) +(1/4) x (24 cm^{2})

= 6 cm^{2 }+ 6 cm^{2}

= 12 cm^{2}

Thus, the required area of parallelogram BDEF = 12 cm^{2}.