Areas of Parallelograms and Triangles Test - 13

Draw a line through D parallel to PQ, which cuts side AC at R. Then,

join PR.

In ΔRDC, P is the mid-point of DC and PQ||DR.

Hence, by converse of mid-point theorem, Q is the mid-point of RC.

∴RQ = QC

i.e., PQ is the median of ΔRPC.

We know that a median of a triangle divides it into two triangles of equal areas.

∴ Area (ΔPQC) = Area (ΔRPQ)

⇒ Area (ΔRPC) = 2area (ΔPQC) … (i)

In ΔRDC, RP is a median. [P is the mid-point of DC]

∴ Area (ΔRPD) = Area (ΔRPC)

⇒ ΔArea (ΔRDC) = 2area (ΔRPC)

= 2 × 2area (ΔPQC) [Using (i)]

= 4area (ΔPQC) … (ii)

In ΔABC, D is the mid-point of BC. [AD is the median with respect to BC]

Also, DR||AB [DR||PQ and PQ||AB]

Hence, by converse of mid-point theorem, R is the mid-point of AC.

i.e., DR is the median of ΔADC with respect to AC.

∴ Area (ΔADC) = 2 (area ΔRDC)

= 2 × 4area (ΔPQC) [Using (ii)]

= 8area (ΔPQC) … (iii)

In ΔABD, AD is the median with respect to BC.

∴ΔArea (ΔABC) = 2area (ΔADC)

= 2 × 8 area (ΔPQC) [Using (iii)]

= 16area (ΔPQC)

= (16 × 6) cm² [Area (ΔPQC) = 6 cm²]

= 96 cm²

The correct answer is D.

It is given that l||m and PB||AC.

Therefore, PACB is a parallelogram.

⇒ PA = BC [In a parallelogram, opposite sides are equal]

Similarly, as l||m, AB||QC, AQCB is a parallelogram.

⇒ AQ = BC

∴PA = AQ

It is seen that ΔPAB and ΔAQC lie between the same parallels and PA = AQ.

We know that two triangles that lie on the same base (or equal bases) and between the same parallels are equal in area.

∴Area (ΔPAB) = Area (ΔAQC)

Thus, the length of AC is 7.5 cm.

The correct answer is C.

Draw a line l passing through A and parallel to side BC.

It is given that in ΔABC, BD = DF = FC.

Therefore, ΔABD, ΔADF, and ΔAFC have equal bases (BD = DF = FC) and they lie between the same parallels.

We know that any two triangles lying on the same base (or equal bases) and between the same parallels are equal in area.

∴Area (ΔABD) = Area (ΔADF) = Area (ΔAFC) = 10 cm²

In ΔADF, DE = EF, i.e., AE is the median of ΔADF with respect to side DF.

∴Area (ΔADE) = Area (ΔAEF)

[Median of a triangle divides it into two triangles of equal areas]

⇒ Area (ΔAEF) =

∴Area (ΔAEC) = Area (ΔAEF) + Area (ΔAFC) = 5 cm² + 10 cm² = 15 cm²

Thus, the area of ΔAEC is 15 cm^2.

The correct answer is B.

In quadrilateral ABCD,

AB||CD [Given]

AD||BC [l||m]

∴ ABCD is a parallelogram.

Similarly, we can show that quadrilaterals PBCQ and PBXY are parallelograms.

BY is a diagonal of parallelogram PBXY.

It is known that the diagonal of a parallelogram divides it into two triangles of equal areas.

∴ Area (PBXY) = 2area (ΔPBY)

= (2 × 15) cm² [Area (ΔPBY) = 15 cm²]

= 30 cm² (i)

It is seen that parallelograms PBXY and PBCQ are lying on the same base PB and between the same parallels PB and n.

∴ Area (PBCQ) = Area (PBXY) (ii)

It is also seen that parallelograms PBCQ and ABCD are lying on the same base BC and between same parallels BC and m.

∴ Area (ABCD) = Area (PBCQ) (iii)

From (i), (ii), and (iii), we obtain

Area (ABCD) = 30 cm²

Thus, the area of quadrilateral ABCD is 30 cm².

The correct answer is C.

In a rhombus, diagonals bisect each other at right angle.

∴ΔOBC is right-angled at point O, where OBand OC

On applying Pythagoras theorem to ΔOBC, we obtain

BC² = OB² + OC²

⇒ BC² = (8 cm)² + (6 cm)²

⇒ BC² = 64 cm² + 36 cm²

⇒ BC² = 100 cm²

⇒ BC = 10 cm

ABCD and PQCB are parallelograms lying on the same base BC and between the same parallels PD and BC.

∴ Area (PQCB) = Area (ABCD)

Area of rhombus ABCD =

Since ABCD is a rectangle, Area (ABCD) = Length × Breadth = BC × PB

∴ BC × PB = 96 cm²

⇒ 10 cm × PB = 96 cm²

Thus, the measure of PB is 9.6 cm.

The correct answer is B.

Draw a parallel through point A to PQ, which intersects PS and QR at X and Y respectively.

Consider quadrilateral PXYQ.

PQ || XY [by construction]

PX || QY [PS || QR]

Hence, PXYQ is a parallelogram.

Parallelograms PXYQ and ΔPAQ are lying on the same base PQ and between the same parallels PQ and XY.

∴ Area (ΔPAQ) =

Similarly, it can be proved that

Area (ΔSAR) =

On adding equations (i) and (ii), we obtain

Area (ΔPAQ) + Area (ΔSAR) =

=$\frac{1}{2}\mathrm{area}\left(\mathrm{PQRS}\right)$

∴Area (ΔPAQ) + Area (ΔSAR) = $\frac{1}{2}\mathrm{area}\left(\mathrm{PQRS}\right) \cdots \left(\mathrm{iii}\right)$

Similarly, by drawing a line passing through B parallel to RS, it can be proved that

Area (ΔPBQ) + Area (ΔSBR) = $\frac{1}{2}\mathrm{area}\left(\mathrm{PQRS}\right) \cdots \left(\mathrm{iv}\right)$

On comparing equations (iii) and (iv), we obtain

Area (ΔPAQ) + Area (ΔSAR) = Area (ΔPBQ) + Area ($∆$SBR)

⇒ Area (ΔSAR) − Area (ΔSBR) = Area (ΔPBQ) − Area (ΔPAQ)

⇒ Area (ASBR) = Area (BPAQ)

⇒ Area (BPAQ) = 35 cm² [Area (ASBR) = 35 cm²]

Area (ΔPAC) + Area (ΔAQD) = Area (BPAQ) − Area (ABCD)

It is given that area (ABCD) = 20 cm²

∴Area (ΔPAC) + Area (ΔAQD) = 35 cm² − 20 cm² = 15 cm²

Thus, area of the shaded portion

= Area (ASBR) + [Area (ΔPAC) + Area ($∆$AQD)]

= 35 cm² + 15 cm²

= 50 cm²

The correct answer is B.

In quadrilateral ABCD,

∠CAD = ∠ACB = 32°

i.e., the pair of alternate interior angles is equal.

∴ AD||BC

It is also seen that ΔABC and ΔDBC lie on the same base BC and between the same parallels and AD and BC.

∴ Area (ΔABC) = Area (ΔDBC)

⇒ Area (ΔAOB) + Area (ΔBOC) = Area (ΔDOC) + Area (ΔBOC)

⇒ Area (ΔAOB) = Area (ΔDOC)

⇒ Area (ΔAOB) = 23 cm² [Area (ΔDOC) = 23 cm²]

Area (ABCD) = Area (ΔAOB) + Area (ΔBOC) + Area (ΔDOC) + Area ΔAOD

⇒ Area (ABCD) = 23 cm² + 53 cm² + 23 cm² + 11 cm² = 110 cm²

The correct answer is B.
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Join BQ and CP

BP is a line parallel to CQ and QP⊥BP, PQ⊥CQ.

∴ ΔPCQ is right-angled at Q.

∴Area (ΔPCQ) =

It is seen that ΔBCQ and ΔPCQ lie on the same base CQ and between the same parallels CQ and BP.

∴Area (ΔBCQ) = Area (ΔPCQ)

⇒ Area (ΔBCQ) = 55 cm²

ΔBDQ and ΔBAD are lying on the same base BD and between the same parallels BD and AQ.

∴ Area (ΔBDQ) = Area (ΔBAD)

⇒ Area (ΔBDQ) + Area (ΔBCD) = Area (ΔBAD) + Area (ΔBCD)

⇒ Area (ΔBCQ) = Area (ABCD)

⇒ Area (ABCD) = 55 cm²

Thus, the area of quadrilateral ABCD is 55 cm².

The correct answer is A.