Mathematics
Class-X
2015-2016
Q.1. Area of two similar triangles ABC and PQR are 25 cm2 and 49 cm2 respectively. If QR
= 9.8 cm, find BC.
Solution :
Given :
Δ ABC ~ Δ PQR Therefore,{ar(Δ ABC)}/{ar(Δ PQR)}= BC2/QR2
or, 25/49 = BC2/(9.8)2 , [Taking square root of both sides]
5/7 = BC/9.8 or, 7BC = 49 or, BC = 7 cm. [Ans.]
Q.2. Two isosceles triangles have equal vertical angles and their areas are in the ratio 4 : 9.
Find the ratio of their corresponding heights.
Solution :
Let ABC and PQR be two isosceles triangles, where, AB = AC, PQ = PR and vertical angles A
and P are equal.
Fig
Also, (area of Δ ABC)/(area of Δ PQR) = 4/9, AD is perpendicular to BC and PM is
perpendicular to QR. As, AB =AC and PQ = PR then AB/PQ = AC/PR ; and vertical angles A
and P are equal ;
Therefore, Δ ABC ~ Δ PQR [S.A.S. similarity axiom]
(area of Δ ABC )/(area of Δ PQR) = AD2/PM2
or, 4/9 = AD2/PM2
or, AD/PM = 2/3. [Ans.]
Q.3. P and Q are points on the side AB and AC respectively of a triangle ABC such that PQ
is parallel to BC and divides triangle ABC into two parts, equal in area. Find PB : AB.
Solution :
Fig
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Mathematics
Class-X
2015-2016
Area of Δ APQ = Area of trapezium PBCQ
= Area of Δ ABC – area of Δ APQ Or, 2×Area of Δ APQ = Area of Δ ABC -----------------------(i)
In Δs APQ and ABC,
L APQ = L ABC [PQ is parallel to BC, corresponding angles are equal]
L A is common , Hence, Δ APQ ~ Δ ABC [A.A. axiom of similarity]
Therefore, area of Δ APQ / area of Δ ABC = AP2/AB2
Or, area of Δ APQ / 2×area of Δ APQ = AP2/AB2
Or, 2AP2 = AB2 Or, √2 AP = AB
Or, √2(AB – PB) = AB Or, (√2 – 1). AB = √2.PB
Hence, PB / AB = (√2 – 1)/√2 Or, PB : AB = (√2 – 1) : √2. [Ans.]
Q.4. In the figure given below, DE is parallel to BC, AD = 4 cm, BD = 2 cm and the area of
Δ ABC = 12 cm2. Calculate :
i.
ii.
Area of Δ ADE
(Area of Δ ADE)/(Area of trapezium DBCE).
Fig
Solution :
i.
As, DE is parallel to BC, therefore, Δ ADE ~ Δ ABC
Therefore, (Area Δ ADE)/(Area of Δ ABC) = AD2/AB2
Or, (Area of Δ ADE)/12 = 42/(4 + 2)2 = 42/62 = 16/36 = 4/9 --- (1)
Or, Area of Δ ADE = 12×(4/9) = 16/3 = 5 1/3 cm2. [Ans.]
ii.
From part (i) we have, (Area of Δ ADE)/(Area of Δ ABC) = 4/9
Or, (Area of Δ ADE)/(Area of Δ ABC – Area of Δ ADE) = 4/(9 – 4)
Or, (Area of Δ ADE)/(Area of trapezium DBCE) = 4/5. [Ans.]
Q.5. In the given figure, ABC is a triangle, DE is parallel to BC and AD/DB = 3/2.
Sudheer Gupta .
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Mathematics
Class-X
2015-2016
Fig
i.
ii.
iii.
Determine the ratio AD/AB, DE/BC.
Prove that ΔDEF is similar to Δ CBF.
Hence, find EF/FB.
What is the ratio of the areas of Δ DEF and Δ BFC ?
Solution :
i.
Given – DE || BC and AD/DB = 3/2.
In ΔS ADE and ABC, LA = LA, (Common Angle)
LD = LB, (Corresponding Angles)
Therefore, Δ ADE ~ Δ ABC [By A A similarity]
Hence, AD/AB = AE/AC = DE/BC
Now, AD/AB = AD/(AD + DB) = 3/(3 + 2) = 3/5 = DE/BC. [Ans.]
ii.
In ΔS DEF and CBF,
LFDE = LFCB (Alternate Angle)
LDFE = LBFC (Vertically Opposite Angle)
Therefore, Δ DEF ~ Δ CBF [By A A Similarity]
Hence, EF/FB = DE/BC = 3/5 => EF/FB = 3/5 [Ans.]
iii.
iv.
Area of Δ DEF / Area of Δ CBF = EF2/FB2 = 32/52 = 9/25 [Ans.]
Q.6. In the figure given below, PB and QA are perpendiculars to the line segment
AB. If PO = 6 cm, QO = 9 cm and the area of ΔPOB = 120 cm2, find the area of
ΔQOA.
v.
vi.
Fig
Solution :
We have, PO = 6 cm, QO = 9 cm, Area of Δ POB = 120 cm2.
In Δs POB and QOB, LB = LA [Each 90º]
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Mathematics
Class-X
vii.
viii.
2015-2016
LPOB = LQOA [Vertically opposite angle]
Therefore, ΔPOB ~ ΔQOA [AA – criteria]
Hence, Area ofΔQOA/Area of Δ POB = OQ2/OP2
Or, Area of ΔQOA/120 = 92/62
Or, Area of Δ QOA = (81×120)/36 = 270 cm2.
Q.7.In the given figure, AB and DE are perpendicular to BC. If AB = 9 cm, DE = 3
cm and AC = 24 cm, calculate AD.
Solution :
ix.
xii.
Fig
In Δs ABC and DEC,
LB = LE = 90º and LC = LC
Therefore, Δ ABC ~ Δ DEC [A.A. similarity rule]
Hence, AB/DE = AC/DC
Or, 9/3 = 24/DC
Or, DC = (24×3)/9 = 8 cm.
Therefore, AD = AC – DC = 24 – 8 = 16 cm. [Ans.]
Q.8. In the given figure, the median BD and CE of a triangle ABC meet at G. Prove
that :
(i) ΔEGD ~ ΔCGB and (ii) BG = 2GD from (i) above.
xiii.
Fig
Solution :
x.
xi.
xiv.
xv.
Fig
We have, BD and CE are median of Δ ABC,
Therefore, AE = EB and AD = DC
Or, AE/EB = AD/DC
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Mathematics
Class-X
xvi.
xvii.
xviii.
2015-2016
Hence, ED || BC.
Now, in Δs EGD and CGB, L1 = L4 [vertically opposite angles are equal]
L3 = L5 [Alternate angles]
L2 = L6 [Alternate angles]
Therefore, ΔEGD ~ Δ CGB [By A.A.A. similarity] [Proved.]
(ii) BD is a median,
Therefore, BG/GD = 2/1 => BG = 2 GD. [Proved.]
Q.9. In the figure alongside, BC || DE. Area of triangle ABC = 25 cm2, area of
trapezium BCED = 24 cm2, DE = 14 cm. Calculate the length of BC.
Fig.
Solution :
In Δs, ABC and ADE, BC || DE,
Therefore, LA = LA [Common]
LB = LD and LC = LE [Corresponding angles]
Therefore, Δ ABC ~ Δ ADE [By A.A.A. similarity]
Therefore, ar(ΔABC)/ar(ΔADE) = BC2/DE2
Or, BC2/(14)2 = ar(ΔABC)/[ar(ΔABC) + ar(trap. BCED)]
Or, BC2/196 = 25/(25 + 24)
Or, BC2 = (25×196)/49 = 100
Or, BC = 10 cm. [Ans.]
Sudheer Gupta .
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Page 5