Mathematics Questions and Answers – Parallelograms & Triangles on the Same Base and Between the Same Parallels – 2

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This set of Mathematics Objective Questions and Answers for Class 9 focuses on “Parallelograms & Triangles on the Same Base and Between the Same Parallels – 2”.

1. Identify the correct relation if PQRS is a parallelogram.

a) ar(POQ) + ar(SOR) > ar(POS) + ar(QOR)
b) ar(POQ) + ar(SOR) = ar(POS) + ar(QOR)
c) ar(POQ) + ar(SOR) < ar(POS) + ar(QOR)
d) ar(POQ) = ar(POS) + ar(SOR)
View Answer

Answer: b
Explanation: Through O, draw AC || PS and BD || PQ

AS POS and PACS are on same base PS and between same parallels PS and AC,
⇒ ar(POS) = ½ ar(PACS)  —————(i)
Similarly, ar(QOR) = ½ ar(QACR)  ————-(ii)
ar(POQ) = ½ ar(PQBD)  ————-(iii)
ar(SOR) = ½ ar(RSBD)  ———-(iv)
Adding equation i and ii, ar(POS) + ar(QOR) = ½ ar(PQRS)  ——-(v)
Adding equation iii and iv, ar(POQ) + ar(SOR) = ½ ar(PQRS)  ——-(vi)
From equation v and vi, ar(POS) + ar(QOR) = ar(POQ) + ar(SOQ).
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2. What is the area of triangle ABD if ar(ABC) = 25cm2 and line l || AB?

a) 25 cm2
b) 45 cm2
c) 30 cm2
d) 60 cm2
View Answer

Answer: a
Explanation: From figure, ABD and ABC are on same base AB and between same parallels. Since areas of triangles on same base and between same parallels are equal, ar(ABD) = ar(ABC) = 25 cm2.

3. What is the ratio of ar(PQR) / ar(QTS) if T and S are midpoints of QR ad PS respectively?

a) 1/2
b) 2
c) 4
d) 1/4
View Answer

Answer: c
Explanation: Since median divides a triangle into 2 triangles of equal area,
ar(ΔPQS) = ½ ar(ΔPQR)  ————(i) (PS is median of ΔPQR)
In ΔQPS, TQ is the median
⇒ ar(ΔQTS) = ½ ar(ΔPQS)
⇒ ar(ΔQTS) = ½ x ½ ar(ΔPQR)  (from equation i)
⇒ ar(ΔQTS) = ¼ ar(ΔPQR)
⇒ ar(ΔPQS) / ar(ΔQTS) = 4.
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4. Which of the following relation is correct if a line through S is drawn parallel to PQ and meets extended QR at A?

a) ar(PQAT) = ar(PQRST)
b) ar(PQAT) > ar(PQRST)
c) ar(PQAT) < ar(PQRST)
d) ar(PQAT) = ½ ar(PQRST)
View Answer

Answer: a
Explanation: Draw TR || PQ and Join AT

Triangle ATR and STR are on same base RT and between same parallels RT and AS
⇒ ar(ΔATR) = ar(ΔSTR)
⇒ ar(ΔATR) + ar(PQRT) = ar(ΔSTR) + ar(PQRT)
⇒ ar(PQAT) = ar(PQRST).

5. Find the ratio of ar(POQ) / ar(PQR) if O is centroid of triangle PQR.

a) 2
b) 1/2
c) 3
d) 1/3
View Answer

Answer: d
Explanation: Centroid is the intersection of medians of a triangle. Hence, PA, QB and RC are the medians.
Since median of triangle divides the triangle into two triangles of equal areas,
ar(PQA) = ar(PRA) = ½ ar(PQR)  ———–(i) (AP is median of triangle PQR)
Similarly, ar(QOA) = ar(AOR) = ½ ar(OQR)  ———–(ii) (OA is the median of triangle QOR)
Subtracting equation ii from i,
ar(PQA) – ar(QOA) = ar(PRA) – ar(AOR)
⇒ ar(ΔPOQ) = ar(ΔPOR)  ————–(iii)
Similarly, ar(QBR) = ar(QBP)and ar(OBR) = ar(OBP)
⇒ ar(QBR) – ar(OBR) = ar(QBP) – ar(OBP)
⇒ ar(ΔPOQ) = ar(ΔQOR)  ————–(iv)
From equation iii and iv, ar(ΔPOQ) = ar(ΔQOR) = ar(ΔPOR)  ————–(v)
Now, ar(ΔPOQ) + ar(ΔQOR) + ar(ΔPOR) = ar(ΔPQR)
⇒ 3ar(ΔPOQ) = ar(ΔPQR)  (from equation v)
⇒ ar(ΔPOQ) = 1/3 ar(ΔPQR).
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6. Find the ratio of ar(TRU) / ar(PQRS) if T and U are midpoints and PQRS is a parallelogram.

a) 3/8
b) 8/3
c) 3
d) 8
View Answer

Answer: a
Explanation: Join QS

In ΔPQS, TU || QS and T and U are midpoint, TU = ½ QS  (by midpoint theorem)
⇒ ar(PTU) = ½ x TU x altitude
⇒ ar(PTU) = ¼ ar(PQS)  ( Since TU = ½ QS and altitude of PTU = ½ altitude of PQS)
⇒ ar(PTU) = 1/8 ar(PQRS)  —————-(i) (Since ar(PQS) = ½ ar(PQRS))
Since UQR and PQRS are on same base QR and between same parallels QR and PS and UQ = ½ PQ,
ar(UQR) = ¼ ar(PQRS)  ———–(ii)
Similarly, STR and PQRS are on same base RS and between same parallels PQ and RS and ST = ½ PS,
ar(STR) = ¼ ar(PQRS)  ————-(iii)
Now, ar(ΔTUR) = ar(PQRS) – ar(PTU) – ar(STR) – ar(UQR)
⇒ ar(ΔTUR) = ar(PQRS) – ¼ ar(PQRS) – ¼ ar(PQRS) – ¼ ar(PQRS)  [From equation i,ii and iii]
⇒ ar(ΔTUR) = 3/8 ar(PQRS).

7. What is the area of quadrilateral ABCD?

a) 100 cm2
b) 102 cm2
c) 56 cm2
d) 85 cm2
View Answer

Answer: b
Explanation: From figure, in ΔDEC, ∠DEC = 90°
⇒ DC2 = DE2 + EC2  (By Pythagoras theorem)
⇒ 132 = DE2 + 52
⇒ DE = 12cm
Now, ar(ABCD) = ar(ABED) + ar(DEC)
⇒ ar(ABCD) = (DE x AD) + (1/2 x DE x EC)
⇒ ar(ABCD) = (12 x 6) + (1/2 x 12 x 5)
⇒ ar(ABCD) = 102 cm2.
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8. Find the area of triangle PQR if PT = PR.

a) 56 cm2
b) 20 cm2
c) 40 cm2
d) 24 cm2
View Answer

Answer: d
Explanation: In right angled triangle, midpoint of hypotenuse is equidistant from the vertices.
⇒ QT = PT = RT = 5 cm
⇒ PR = PT + RT = 10 cm
Now, PT2 = PQ2 + QR2  (By Pythagoras theorem)
⇒ 102 = 82 + QR2
⇒ QR = 6 cm
Now, ar(PQR) = ½ x PQ x QR = ½ x 8 x 6 = 25 cm2.

9. If rectangle ORST is inscribed in quadrant of circle of radius 13cm and OT = 5cm, what is the are of quadrilateral ORST?

a) 180 cm2
b) 60 cm2
c) 100 cm2
d) 25 cm2
View Answer

Answer: b
Explanation: From figure, OS = radius of quadrant POQ = 13 cm
and OT = 5 cm
In ΔSOT, OS2 = OT2 + TS2  (By Pythagoras theorem)
⇒ 132 = 52 + TS2
⇒ TS = 12 cm
Now, ar(ORST) = OT x TS = 12 x 5 = 60 cm2.
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10. Find the area of quadrilateral PQRS.

a) 120 cm2
b) 76 cm2
c) 60 cm2
d) 150 cm2
View Answer

Answer: a
Explanation: From figure, in ΔPTS, PS2 = PT2 + TS2  (By Pythagoras theorem)
⇒ 102 = PT2 + 62
⇒ PT = 8 cm
Similarly, UQ = 8cm
Since UTRS is rectangle, RS = UT = 12 cm
⇒ PQ = PT + UT + UQ = 28 cm
Now, ar(trapezium PQRS) = ½ x sum of parallel sides x altitude
⇒ ar(PQRS) = ½ x (28 + 12) x 6 = 120 cm2.

Sanfoundry Global Education & Learning Series – Mathematics – Class 9.

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Manish Bhojasia, a technology veteran with 20+ years @ Cisco & Wipro, is Founder and CTO at Sanfoundry. He is Linux Kernel Developer & SAN Architect and is passionate about competency developments in these areas. He lives in Bangalore and delivers focused training sessions to IT professionals in Linux Kernel, Linux Debugging, Linux Device Drivers, Linux Networking, Linux Storage, Advanced C Programming, SAN Storage Technologies, SCSI Internals & Storage Protocols such as iSCSI & Fiber Channel. Stay connected with him @ LinkedIn | Youtube | Instagram | Facebook | Twitter