| Exercise-I |
| 1. |
If AD = 12 cm , BD = 9 cm, AD  BC and  = 90 then the area of the triangle is |
| Sol: |
AD BC |
| |
As Δ ADB ~ Δ ADC |
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AD2 |
= |
BD × DC |
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122 |
= |
9 × DC |
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CD |
= |
 |
| |
|
= |
16 cm |
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BC |
= |
BD + CD |
| |
|
= |
9 + 16 |
| |
|
= |
25cm |
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Area |
= |
 |
| |
|
= |
 |
| |
|
= |
150cm2 |
|
|
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| 2. |
D, E, F are midpoints of the sides BC, AC and AB respectively. BX DF. If the area of the ΔABC is 24cm2 then the length of DF × BX is |
| Sol: |
Area of Δ BFD |
= |
 |
| |
|
= |
 |
| |
|
= |
6cm2 |
| |
Area of Δ BFD |
| |
 |
= |
6 cm2 |
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DF × BX |
= |
6 × 2 cm2 |
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DF × BX |
= |
12 cm2 |
|
|
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| 3. |
Δ DEC is an equilateral triangle, ABCD is a square. If DM EC and DM = 24cm, what is the length of the diagonal of the square? |
| Sol: |
DM |
= |
Altitude of the Δ DEC |
| |
DM |
= |
 |
| |
24 |
= |
|
| |
side of Δ DEC : CD |
= |
 |
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side of square ABCD |
= |
|
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Diagonal of the square ABCD |
= |
√2 × side |
| |
|
= |
 |
| |
|
= |
 |
| |
|
= |
 |
| |
|
= |
 |
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| 4. |
The value of the nth term of a sequence is given by the expression 3n – 7. If the 10th term is a and 15th term is b then a – b is |
| Sol: |
tn |
= |
3n – 7 |
| |
b |
= |
t15 |
| |
|
= |
3 × 15 – 7 |
| |
|
= |
45 – 7 |
| |
|
= |
38 |
| |
a |
= |
t10 |
| |
|
= |
3 × 10 – 7 |
| |
|
= |
30 – 7 |
| |
|
= |
23 |
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t10 – t15 |
= |
a – b |
| |
|
= |
23 – 38 |
| |
|
= |
– 15 |
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|
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| 5. |
Number of rectangles in the given figure
 |
| Sol: |
Number of Rectangles |
= |
9 |
| |
They are AEIH, EFJI, FBGK, KGCL, HJLD, AFJH, FBCL, AFLD, ABCD |
|
|
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| 6. |
Count the number of triangles in the following figure
 |
| Sol: |
The number of triangles are 17 |
| |
They are AFC, AFB, BGF, CGF, CGE, BGD, EHG, DHG, ABC, ACG, GCB, ABG, GDE, BCD, CEB, EDC, EDB. |
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| 7. |
In quadrilateral ABCD,  = 90 , BC = 38cm and DC = 25cm. A circle is inscribed in this quadrilateral which touches AB at point Q such that QB = 27cm. Find the radius of the circle if BQ = 27cm |
| Sol: |
BQ = BR = 27cm |
| |
Let x be the radius of the circle |
| |
OP = OS = PD = x |
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CR |
= |
BC – BR |
| |
|
= |
38 – 27 |
| |
|
= |
11cm |
| |
Tangents from external points are equal |
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DS |
= |
CD – CS |
| |
|
= |
25 – 11 |
| |
|
= |
14 cm |
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Both DS = OP = x |
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x |
= |
14 |
| |
Radius of circle = 14cm |
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|
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| 8. |
There is an incircle in a right angled PQR. Find the radius of the circle. |
| Sol: |
Area |
= |
 |
| |
Let the radius of circle be r |
| |
Area of ΔPQR |
= |
Area of Δ POQ + Δ QOR + Δ POR |
| |
 |
= |
 |
| |
30 |
= |
 |
| |
60 |
= |
30r |
| |
r |
= |
2cm |
|
|
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| 9. |
A chord of length 16cm is at a distance of 15cm from the centre of the circle.Find the length of the chord of the same circle which is at a distance of 8cm from the centre
 |
| Sol: |
By pythogoras theorem |
| |
AC2 + OC2 |
= |
OA2 |
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Where AC |
= |
 |
| |
|
= |
8cm |
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82 + 152 |
= |
OA2 |
| |
OA2 |
= |
289 |
| |
OA |
= |
 |
| |
OA |
= |
17 cm (radius of the circle) |
| |
 OX |
= |
17cm |
| |
OX2 |
= |
OP2 + PX2 |
| |
172 |
= |
82 + PX2 |
| |
PX2 |
= |
289 – 64 |
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PX |
= |
 |
| |
|
= |
15 cm |
| |
PX = PY = 15cm |
| |
XY |
= |
PX + PY |
| |
XY |
= |
15 + 15 |
| |
|
= |
30cm |
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