CBSE Class 10 Mathematics Triangles Worksheet Set 03

Question. It is given that ΔABC ~ ΔPQR with BC/QR = 1/3 , then find the value of ar(ΔPRQ) / ar(ΔACB).
Ans. 1/9

Question. In the given figure, if ΔABC ~ ΔPQR, find the value of x?

Ans. AB/PQ = BC/QR ⇒ 6/4.5 = 4/x ⇒ x = 3cm

Question. In the given figure, find the value of x which will make DE || AB ?

Ans. (x+3) / (3x+19) = x/(3x+4)  (By B.P.T.)
x = 2

Question. An aeroplane leaves an airport and flies due west at a speed of 2100 km/hr. At the same time, another aeroplane leaves the same place at airport and flies due south at a speed of 2000 km/hr. How far apart will be the two planes after 1 hour?
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Question. If ΔABC and ΔDEF are similar triangles such that ∠A = 45° and ∠F = 56°, then find the value ∠C.
Ans. ∠F = ∠C = 56°

Question. In the given figure, XY || QR and PX/XQ = PY/YR = 1/2 , find XY : QR.

Ans. ΔPXY ~ ΔPQR
PX/PQ = XY/QR = 1/3
∴ XY : QR = 1 : 3

Question. In the given figure, AB || DE. Find the length of CD.

Ans. ΔABC ~ ΔEDC            (AA Similarity criterion)
6/3 = 5/CD
CD = 2.5 cm

Question. Perimeter of two equilateral triangles ΔABC and ΔPQR are 144 m and 96 m, Find ar (ΔABC) : ar (ΔPQR).
Ans. ar(ΔABC)/ar(ΔPQR) = (144/96)² = 9/4
∴ ar(ΔABC) : ar(ΔPQR) = 9 : 4

Question. In the given figure, ABCD is a parallelogram. AE divides the line segment BD in the ratio 1 : 2. If BE = 1.5 cm find BC.

Ans. ΔBOE ~ ΔDOA          (AA Similarity criterion)
BO/DO = BE/DA
1/2 = 1.5/DA
DA = 3 cm
BC = DA = 3 cm              (Opposite sides of a parallelogram)

Question. In ΔABC, AD ⊥ BC such that AD² = BD × CD. Prove that ΔABC is right angled triangle.
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Question. In figure ΔPQR is right-angled at P.M is a point on QR such that PM is perpendicular to QR. Show that PQ² = QM × QR.

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Question. Through the vertex D of a parallelogram ABCD, a line is drawn to intersect the sides BA and BC produced at E and F respectively. Prove that DA/AE = FB/BE = FC/CD.
Ans. 

ΔEAD ~ ΔEBF
EA/EB = AD/BF
⇒ BF/BE = AD/AE = (BF - AD)/(BE - AE) = (BF - BC)/BA = CF/DC

Question. In figure BN and CM are medians of a ΔABC right angled at A. Prove that 4(BN² + CM²) = 5 BC²

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Question. In equilateral ΔABC, AD ⊥ BC. Prove that 3BC² = 4AD².
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ΔADB ≅ ΔADC
BD = DC
∴ BD = (1/2)BC       ..(1)
In right angled ΔADB,
AB² = AD² + BD²
BC² = AD² +(BC/2)² [∵ AB = BC = CA and from (1)]
3BC² = 4AD²

Question. The Perimeters of two similar triangles are 30 cm and 20 cm repsectively. If one side of the first triangle is 9 cm long. Find the length of the corresponding side of the second triangle.
Ans. ΔABC ~ ΔDEF
∴ AB/DE = BC/EF = AC/DF = k
⇒ AB = kDE, BC = kEF, AC = kDF
∴ AB + BC + AC = k (DE + EF + DF)
∴ 30/20 = 9/x ⇒ x = 6 cm

Question. In the given figure, the line segment XY is Parallel to AC of ΔABC and it divides the triangle into two parts of equal areas. Prove that AX/AB = (√2-1) / √2

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Question. Sides AB and AC and median AD of ΔABC are respectively proportional to sides PQ and PR and median PM of ΔPQR. Show that ΔABC ~ ΔPQR.
Ans. In ΔABC and ΔPQR
AB/PQ = AC/PR = AD/PM               ...(1)
Extend AD to a point E s.t. AD = DE and PM to point L s.t. PM = ML

∴ quadrilateral of ABEC and PQLR are parallelogram
(∵ diagonals bisect each other)
Δ AC = BE, AB = EC
PR = QL, PQ = LR                      ....(2)
From (1) and (2)
AB/PQ = BE/QL = 2AD/2PM = AE/PL
∴ ΔABE ~ ΔPQL
∴ ∠ABE = ∠PQL                           ...(3)
Similarly, ΔAEC ~ ∠PLR
⇒ ∠CAE = ∠RPL                          ...(4)
⇒ ∠CAB = ∠RPQ                (from 3 and 4)
∴ In ΔABC and ΔPQR
AB/PQ = AC/PR and ∠CAB = ∠RPQ
∴  ΔABC ~ ΔPQR

Question. In figure if ΔABC ~ ΔDEF and their sides of lengths (in cm) are marked along them, then find the lengths of sides of each triangle.

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