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⚫ Level – 2 & 3  Questions: 25

 EXERCISE – 12  Time: 25 Min.
1. A, B, C and D are the vertex of rhombus. Side AB and BA (a) πr2 (b) 2πr2 (c) 1.5πr2 (d) 1.25πr2
are extended to P and E respectively. Extended part of 5. In ABC, AM is the angle bisector of ∠BAC, AB + AC =
ED and PC meets at point T. If EA = AB = BP. Find ∠T.
7.2 cm, BM = 2.5 cm and MC = 2. Find AB
A, B, C ,oa D leprqHkqt
Z ds fcUnq gSA Hkqtk AB ,oa BA dks
f=Hkqt ABC es]a ∠BAC dk dks.k lef}Hkktd AM gS] AB+AC =
Øe'k% P ,oa E fcUnq rd vkxs c<+k;kA ED o PC dk c<+k gqvk
7.2cm, BM = 2.5cm ,oa MC = 2. AB Kkr djks&
Hkkx fcUnq T ij feyrk gSA vxj EA = AB = BP gS rks ∠T dk
eku Kkr djsAa
(a) 600 (b) 300 (c) 1500 (d) 900

2. In the given figure AD = DE = EC = BC, AB = AC find ∠A

fn, x;s fp= esa AD = DE = EC = BC, AB = AC ∠A dk eku (a) 3.2 (b) 4 (c) 5 (d) 2.2

Kkr djsAa 6. Find the maximum area that can be enclosed in a

triangle of perimeter 12 cm.

og vf/kdre {kS=Qy Kkr dhft, tks 12 lseh ifjfefr okys

f=Hkqt ls f?kjk gksA

(a) 4 cm² (b) cm² (c) cm² (d) 2 cm²

180 180 180 180
(a) (b) (c) (d)
11 13 7 17 7. In the figure AB = 8 cm, BC = 7cm, ∠ ABC = 1200 Find AC.
3. In a triangle ABC, right angle at B. AX = AD, CY = CD. Find fuEu vkd`fr esa AB = 8cm, BC = 7cm, ∠ABC = 120°, AC Kkr
∠XDY. djksAa
f=Hkqt ABC es]a B ij ledks.k gS] AX = AD, CY = CD. ∠XDY
Kkr dhft,A

(a) 11 (b) 12 (c) 13 (d) 14

(a) 300 (b) 450 (c) 600 (d) 900 8. PQRS is a circle and circles are drawn with PO, QO, RO &
SO as diameter as shown in the figure. A/B is equal to
4. PQRST is a pentagon in which all the interior angles are
unequal. A circle of radius 'r' is inscribed in each of the PQRS ,d o`Ùk gS ,oa mlds vUnj PO, QO, RO ,oa SO O;kl
vertices. Find the area of portion of circles falling inside okys o`Ùk cus gS tSlk fd vkd`fr esa fn[kk;k x;k gSA A/B
the pentagon. cjkcj gksxk&
PQRST ,d iapHkqt gS ftlds lHkh vUr% dks.k cjkcj ugh gSA
'r' f=T;k okys o`Ùk lHkh fcUnqvksa ij cus gq, gSA o`Ùkksa ds os
Hkkx tks iapHkqt ds vanj vkrs gSa dk {kS=Qy Kkr dhft,A

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Q
,d f=Hkqt esa nks Å¡pkbZ 4 lseh o 12 lseh gS] rhljh Å¡pkbZ
A A dh vf/kdre Å¡pkbZ fdruh gks ldrh gS tc lHkh Å¡pkbZ
B B
P R
,d iw.kkZad la[;k gksA
B B
(a) 4 (b) 5 (c) 6 (d) 7
A A

S
13. If AB = BC = CD = DE = EA = AF = FB, then find .

(a) π (b) 1 (c) π/4 (d) 2 ;fn AB = BC = CD = EA = AF = FB,  Kkr djksA

9. Two posts of heights 6m and 11m stand on a plane
ground. If the distance between the feets of posts is 12
m, find the distance between their tops.

,d lery ij 6 ehVj o 11 ehVj Å¡pkbZ okys nks iksLV

(a) 840 (b) 1020 (c) 900 (d) 960
[kM+s gSA ;fn nksuksa iksLVksa ds chp dh nwjh 12 ehVj gS rks
14. ‘D’ is a point on side ‘BC’ of a triangle ABC. Such that AD
nksuks ds f'k[kjksa ds chp dh nwjh gksxh&
= 13 cm, BC = 21 cm, CD = 14 cm and AC = 15 cm. The
(a) 10 (b) 11 (c) 12 (d) 13
area of triangle ABC is.
10. If the measures of sides of triangle are (x²-1), (x²+1) and
ABC f=Hkqt dh Hkqtk BC ij ,d fcUnq D gSA bl izdkj AD =
2x cm, then the triangle would be ?
13, BC = 21cm, CD = 14cm o AC = 15 cm f=Hkqt ABC dk
;fn f=Hkqtks dh Hkqtkvksa dk eki (x²-1), (x²+1) o 2x gS rc
{kS=Qy Kkr djksA
f=Hkqt gksxk\
(a) 125 (b) 126 (c) 127 (d) 128
(a) Equilateral (b) Right angle
15. In the given figure AE : EC = 3 : 2 AF : FD = 9 : 4 then BD : DC is
(c) obtuse angle (d) acute angle
fuEu vkd`fr esa AE : EC = 3 : 2, AF : FD = 9 : 4 rc BD : DC gSA
11. In ABC, AB = 13 cm, BC = 14 cm and AC = 15 cm. AD is

an altitude and line segment EF is Perpendicular to BC.

EF divides ABC into two equal areas. What is the area

of trapezium ADFE.
(a) 1 : 2 (b) 2 : 1 (c) 3 : 1 (d) 1 : 3
f=Hkqt ABC esa] AB = 13cm, BC = 14cm o AC = 15cm, AD
16. In the figure AE : EB = AF : FC = 1 : 4, DG : GB = DH : HC =
Å¡pkbZ gS rFkk EF, BC ij yEcor~ gSA EF, ∆ABC dks nks leku
1 : 3, if EF = 4 cm, then GH = ?
{kS=Qyksa esa foHkkftr djrk gSA lery prqHkqt
Z ADFE dk
vkd`fr esa AE : EB = AF : FC = 1 : 4, DG : GB = DH : HC = 1 :
{kS=Qy D;k gksxk\
3, ;fn EF = 4 cm, gks rks GH = ?

(a) 10 (b) 15 (c) 18 (d) 12

12. In a triangle two altitudes are 4 cm, 12 cm respectively,

and all the altitudes are integers, what could be the (a) 3 cm (b) 4 cm (c) 5 cm (d) 6 cm
largest possible length of the third altitude 17. Find the value of ‘x’, where it is known that ‘x’ is a
natural number and C and E are right angle

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x dk eku Kkr djks] tgkW x ,d izkd`r la[;k gS ,oa ∠C o 22. ABCD is a rectangle with sides AB = 26cm and BC =

∠E ledks.k gSA 22cm. Four congruent triangles are cut off from each the
corners a rectangle ABCD to get a regular octagon. What
A B
is the length of the side of octagon?

ABCD ,d vk;r gS ftldh Hkqtk,W AB = 26cm o BC =

E
22cm gSA ABCD vk;r ds pkjksa dks.kksa ls pkj lokZaxle
D C
f=Hkqtksa esa dkVk tkrk gS rkfd lev"VHkqt izkIr gksA v"VHkqt
(a) 9 (b) 10 (c) 11 (d) 12 ds Hkqtk dh yEckbZ D;k gksxh\
18. If ABC is a right angle triangle with hypotenuse AC = 15 (a) 8 (b) 10 (c) 12 (d) 14
cm, points M and N trisect the side AC, then BM² + BN² = ? 23. In the given figure ABCD is a square of side 12cm.A line
;fn ∆ABC ,d ledks.k f=Hkqt gS ftldk d.kZ AC = 15cm gSA CE is drawn as shown in figure. If radius of smaller circle
fcUnq M o N Hkqtk AC dks lef}Hkkftr djrs gS rc BM² + BN² = ? is 3cm then find the radius of bigger circle.

ABCD ,d oxZ gS ftldh Hkqtk 12 lseh gS ,d js[kk CE [khaph

tkrh gS tSlk vkd`fr esa fn;k gSA NksVs o`Ùk dh f=T;k 3 lseh
gS rks cM+s o`Ùk dh f=T;k Kkr djksA

(a) 120 (b) 130 (c) 115 (d) 125

19. In the given figure AC = BC, FD and FE angle bisector. If

angle CAB = 500 then ∠DFE
(a) 3.5cm (b) 4cm (c) 5cm (d) 4.5cm
nh x;h vkd`fr esa AC = BC, FD o FC dks.k lef}Hkktd gSA
24 In the given figure AE: EB = 3: 2, AF: FC = 1: 5 and BD:
;fn ∠CAB = 50° rc ∠DFE
CD = 4: 3, then find the ratio of areas of EFCD to Δ ABC.

nh x;h vkd`fr esa AE : EB = 3 : 2, AF : FC = 1 : 5 o BD : CD

= 4 : 3 rc EFCD ls ∆ABC ds {kS=Qyksa dk vuqikr Kkr djksa\

(a) 120 (b) 130 (c) 140 (d) 150

20. Two of the sides of a scalene triangle are 10 and 33 cm.

How many different integral values third side can take

,d fo"keckgq f=Hkqt dh nks Hkqtk,W 10 lseh o 33 lseh gSA (a) 45: 71 (b)54:79 (c) 47:70 (d)50: 81

rhljh Hkqtk ds fy, fdrus iw.kkZad eku ys ldrs gS\ 25. In an equilateral triangle ABC, AC ⏊BP and BC⏊PQ.

(a) 18 (b) 19 (c) 20 (d) 21 Find the ratio of area of PRB and PCQ if RP||BC.

21. Find Area of given figure leckgq f=Hkqt ABC es]a AC⊥BP o BC⊥PQ. ;fn RP||BC gks

nh x;h vkd`fr dk {kS=Qy Kkr dhft,A rks ∆PRB o ∆PCQ dk {kS=Qy Kkr djksAa

(a) 48 (b) 56 (c) 24 (d) 96 (a) 2: 1 (b) 5: 2 (c) 3:1 (d)4:3

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Answer Key Assume ∠A = 600

1 d 2 c 3 b 4 c 5 b ∴∠ADC = ∠AXD = 600; ∠C = 300
6 a 7 c 8 b 9 d 10 b
∠YDC = = 750
11 d 12 b 13 d 14 b 15 b
16 c 17 c 18 d 19 b 20 a ∴∠ XDY = 180 – (60 + 75) = 450
21 a 22 b 23 b 24 c 25 a

Solutions 4.(c)
Q
1.(d) P

T R

∵ 2 + 2x = 1800

 + x = 900 (Interior angles of parallel S

sides) As we know that sum of all the interior angles of a
pentagon is 540°.
In TEP
540
∠T + ∠E + ∠P = 1800 ∴ Area of circle inside pentagon = πr2
360
∠ T + + x = 1800 ∠ T = 900
Area = 1.5πr2
Neon Approach :
Assume ABCD is a square 5.(b) Let AB = x ⇒AC = 7.2 – x
∴ So ∠ DEA = 450, ∠CPB = 450
then =
∴∠ T = 900
 =  =  x = AB = 4 cm
2.(c)
reduce calculation :
AB : AC = 2.5 : 2

2.5 + 2 = 4.5 7.2 1

2.5 × 2.5 = × 2.5 = 4 cm

In ABC

x + 3x + 3x = 1800 x=
6.(a) For maximum area, triangle should be equilateral.
3.(b) ∠XAD = 180 - 2 ⇒∠BCA = 2 - 90 ∴ side = = 4 cm

∴∠YDC = DYC = = 135 –

area = (4)² = ×4×4=4 cm²

7.(c) ∠ ABM = 600

∴ AB = 8 cm, AM = AB = 4 cm  MB = = 4 cm

AC² = (AM)² + (MC)² = (4 )² + (4 + 7)²

∴∠ ADX + ∠XDY + ∠YDC = 1800 = 48 + 121 = 169  AC = = 13 cm
 + ∠XDY + 135 – = 1800  = 450
Neon Approach : The only hypotenuse from the
Neon Approach :- option in triplet is 13. So ans is 13.

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12.(b) <h< ⇒h < ⇒h < ⇒ h < 6, So h = 5

8.(b) In the given figure let us find the value of A first. Let
2x be the radius of bigger circle and x be the radius
of all four smaller square. 13.(d) As AB = BC = CD = DE = EA
4A = π(2x)2 - 4π x2 + 4B  4A = 4πx2 - 4πx2 + 4B ∴ ABCDE is a regular pentagon
A ∴ internal angle of regular pentagon
4A = 4B  =1
B
= = × 180 = = 1080

9.(d) ∴∠ ABC = 1080

∵ FBA is a equilateral .

∠ FBA = 600
∴∠FBC = 108 – 60 = 480

In ACB, ∠ACB = ∠CAB = = = 36

∴ = 180 – (48 + 36) = 180 – 84 = 960

DE = BC = 12

∴ AC = = = 13 cm 14.(b) ∵ BD = 21 – 14 = 7 cm
area of ADC = 84 cm²
10.(b) As we can see that (x²+1) will be the largest side [By heron’s formulae]

So Cos = =0

∴ =

∴ = 90

∴ triangle is right angle triangle

11.(d) Semi perimeter of ABC = = 21 cm ∴ ADC = = 42 cm²

∴ Area = = 84 cm² ∴ ABC = 42 + 84 = 126 cm²

15.(b) Draw a line parallel to EF from D.

Assume length of AC = 15 cm

By triplets
AD = 12 cm ⇒BD = 5 cm
∴ area of ∆ABD = × AD × BD = × 12 × 5 = 30 cm² AE = × 15 = 9 cm, EC = 6 cm

area of AEFB = = 42 cm² ∵ EF || DG

∴ area of AEFD = 42 – 30 = 12 cm² =  =  GE = 4 cm

∴ GC = 6 – 4 = 2 cm
BD : DC = EG : GC = 4 : 2 = 2 : 1

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⟹
16.(c) EF = 4 cm then BC = 4 (1+4) = 20 cm
⟹ ⟹
∴ GH = = 5 cm

⟹ ⟹
17.(c) Difference of sides = x + 4 – (x-3) = 7 cm
So, side of octagon = 2a = 2.5 = 10cm
In triplets 7 is the difference when 8, 15, 17
So x – 3 = 8
23.(b)
 x = 11
On putting this value in diagram it is satisfying

18.(d) ∴ BM² + BN² = 8² + 6² + 4² + 3²

= 64 + 36 + 16 + 9 = 125 cm
∵ΔDCE IBC

19.(b) ∠ CAB = ∠ABC = 500 ∴ ⟹ ⟹ = 4cm

∠ACB = 180 – 100 = 800 Alternative

∠D + ∠ E = 180 – 80 = 100
In AFD, radius is 3cm.
∴∠FDE + ∠FED = = 500
12 + n − 144 + n²
=3
∠DFE = 180 – 50 = 1300 2
(6+n)² = 144+n² ⇒ n = 9
20.(a) 33 – 10 ∠ a ∠ 33 + 10 Hence FD = 15 and FB = 3 cm.
23 ∠ a ∠ 43 Now EFB & EDC are similar.
So total values = 19 but is scalene triangle So side BF 3
= ∴ EB = 4 , EF = 5
can’t be 33. So total values possible are 19 – 1 = 18 CD 12
Hence in radius of EDC
21.(a) ∴ area = = 48 12 + 16 − 20
= 4cm = r
2

24.(c)

22.(b)
25.(a)

Let the side of regular octagon = 2a

Suppose ABC is an equilateral triangle of side 6cm
∴ ID = and P is midpoint of AC.

DJ = RP = 3, QC = 3cosC = 3cos60 =

In Δ IDJ

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