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Science & Math for K6 - K12 Students

Geometry SRQ
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 AD2 = BD × DC 122 = 9 × DC CD = = 16 cm BC = BD + CD = 9 + 16 = 25cm Area = = = 150cm2

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 DF × BX = 6 × 2 cm2 DF × BX = 12 cm2

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 = side of square ABCD = Diagonal of the square ABCD = √2 × side = = = =

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 t10 – t15 = a – b = 23 – 38 = – 15

5. Number of rectangles in the given figure
 Sol: Number of Rectangles = 9 They are AEIH, EFJI, FBGK, KGCL, HJLD, AFJH, FBCL, AFLD, ABCD

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.

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 CR = BC – BR = 38 – 27 = 11cm Tangents from external points are equal DS = CD – CS = 25 – 11 = 14 cm Both DS = OP = x x = 14 Radius of circle = 14cm

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

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 Where AC = = 8cm 82 + 152 = OA2 OA2 = 289 OA = OA = 17 cm (radius of the circle) OX = 17cm OX2 = OP2 + PX2 172 = 82 + PX2 PX2 = 289 – 64 PX = = 15 cm PX = PY = 15cm XY = PX + PY XY = 15 + 15 = 30cm

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