Given: points P, Q and R lies on circle c (O, r). Construction: > join PR and QR. > Draw the perpendicular bisectors of PR and QR which intersects at point O. > Taking O as centre and OP as radius, draw a circle. > This is the required circle.
Given: points P, Q and R lies on circle c (O, r).
Construction:
> join PR and QR.
> Draw the perpendicular bisectors of PR and QR which intersects at point O.
> Taking O as centre and OP as radius, draw a circle.
> This is the required circle.
Steps of construction (i) Draw a ray AB at the point A. (ii) Taking A as centre and a convenient radius, draw an arc which intersect AB at C. (iii) Taking C as center and with the same radius, draw an arc which intersect the previous arc at E. (iv) Similarly, taking E as center and with the same radRead more
Steps of construction
(i) Draw a ray AB at the point A.
(ii) Taking A as centre and a convenient radius, draw an arc which intersect AB at C.
(iii) Taking C as center and with the same radius, draw an arc which intersect the previous arc at E.
(iv) Similarly, taking E as center and with the same radius, draw an arc which intersect at F.
(v) Taking E and F as centre, draw arcs with equal radius (more than half of EF), which intersect at H.
(vi) Draw a ray AG. ∠PAQ is the required angle of 90°.
Justification: Join AE, CE, EF, FG, and GE.
AC = CB = AE [∵ By Construction]
⇒ ΔACE is an equilateral Triangle.
⇒ ∠CAE = 60 …(1)
Similarly, ∠AEF = 60° …(2)
Hence, ∠CAE = ∠AEF [∵ from (1) and (2)]
∠CAE and ∠AEF alternate angles, therefore
FE ∥ AC
Here, FG = EG [∵ By Construction]
⇒ Point G lies on the perpendicular bisector of EF. ⇒ ∠GIE = 90°
Hence, ∠GAB = ∠GIE = 90° [∵ corresponding angles]
Steps of construction (i) Draw a ray AB at the point A. (ii) Taking A as centre and a convenient radius, draw an arc which intersect AB at C. (iii) Taking C as centre and with the same radius, draw an arc which intersect the previous arc at D. (iv) Similarly, taking D as centre, draw arcs with the sRead more
Steps of construction
(i) Draw a ray AB at the point A.
(ii) Taking A as centre and a convenient radius, draw an arc which intersect AB at C.
(iii) Taking C as centre and with the same radius, draw an arc which intersect the previous arc at D.
(iv) Similarly, taking D as centre, draw arcs with the same radius, draw an arc which intersects at E.
(v) taking D and E as centre, draw arcs with equal radius (more than half of DE), Which intersect at F.
(vi) From the point A, draw a ray AF, which intersects arc DE at G.
(vii) Taking C and G as centre. Draw arcs with equal radius (more than half of CG), which intersect at H.
(viii) From the point A, draw a ray of 45°.
Justification: Join GH and CH.
In ΔAGH and ΔACH,
GH = CH [∵ Arcs of equal radii]
AG = AC [∵ Radii of same circle]
AH = AH [∵ Common]
So, ΔAGH ≅ ΔACH [∵ SSS Congruency rule]
∠GAH = ∠CAH [∵ CPCT]
Hence, ∠GAH = ∠CAH = 45°
Here, the sides of triangle ABC are a = 28 cm, b = 26 and c = 30 cm. So, the semi-perimeter of triangle S = a+b+c/2 = 28+26+30+ = 84/2 = 42 cm Therefore, using Heron's formula, area of triangle = √s(s - a)(s -b)(s -c) = √42(42 -28)(42 -26)(42 -30) = √42(14)(16)(12) = √112896 = 366 cm² We know that tRead more
Here, the sides of triangle ABC are a = 28 cm, b = 26 and c = 30 cm.
So, the semi-perimeter of triangle S = a+b+c/2 = 28+26+30+ = 84/2 = 42 cm
Therefore, using Heron’s formula, area of triangle = √s(s – a)(s -b)(s -c)
= √42(42 -28)(42 -26)(42 -30) = √42(14)(16)(12) = √112896
= 366 cm²
We know that the area of a parallelogram = base × corresponting hieght
According to question:
Area of parallelogram = Area of triangle
⇒ base × corresponding height = 336
⇒ 28 × corresponding height = 336
⇒ corresponding height = 336/28 = 12 cm.
Join the diagonal AC of quadrilateral ABCD. Here, the sides of triangle ABC are a = 30 m, b = 30 m and c = 48 m. So, the semi-perimeter of triangle S = a + b + c/2 = 30 + 30 + 48/2 = 108/2 = 54 m. Therefore, area of triangle = √s(s - a)(s -b)(s -c) = √54(54 -30)(54 -30)(54 -48) = √54(24)(24)(6) = √1Read more
Join the diagonal AC of quadrilateral ABCD.
Here, the sides of triangle ABC are a = 30 m, b = 30 m and c = 48 m.
So, the semi-perimeter of triangle
S = a + b + c/2 = 30 + 30 + 48/2 = 108/2 = 54 m.
Therefore, area of triangle = √s(s – a)(s -b)(s -c)
= √54(54 -30)(54 -30)(54 -48) = √54(24)(24)(6) = √186624 = 432 m²
Hence, area of quadrilateral = 2 × 432 = 864 m²
Therefore, the area grazed by each cow = Total area/number of cows = 864/18 = 48 m²
Suppose you are given a circle. Give a construction to find its center.
Given: points P, Q and R lies on circle c (O, r). Construction: > join PR and QR. > Draw the perpendicular bisectors of PR and QR which intersects at point O. > Taking O as centre and OP as radius, draw a circle. > This is the required circle.
Given: points P, Q and R lies on circle c (O, r).
See lessConstruction:
> join PR and QR.
> Draw the perpendicular bisectors of PR and QR which intersects at point O.
> Taking O as centre and OP as radius, draw a circle.
> This is the required circle.
Construct an angle of 90° at the initial point of a given ray and justify the construction.
Steps of construction (i) Draw a ray AB at the point A. (ii) Taking A as centre and a convenient radius, draw an arc which intersect AB at C. (iii) Taking C as center and with the same radius, draw an arc which intersect the previous arc at E. (iv) Similarly, taking E as center and with the same radRead more
Steps of construction
See less(i) Draw a ray AB at the point A.
(ii) Taking A as centre and a convenient radius, draw an arc which intersect AB at C.
(iii) Taking C as center and with the same radius, draw an arc which intersect the previous arc at E.
(iv) Similarly, taking E as center and with the same radius, draw an arc which intersect at F.
(v) Taking E and F as centre, draw arcs with equal radius (more than half of EF), which intersect at H.
(vi) Draw a ray AG. ∠PAQ is the required angle of 90°.
Justification: Join AE, CE, EF, FG, and GE.
AC = CB = AE [∵ By Construction]
⇒ ΔACE is an equilateral Triangle.
⇒ ∠CAE = 60 …(1)
Similarly, ∠AEF = 60° …(2)
Hence, ∠CAE = ∠AEF [∵ from (1) and (2)]
∠CAE and ∠AEF alternate angles, therefore
FE ∥ AC
Here, FG = EG [∵ By Construction]
⇒ Point G lies on the perpendicular bisector of EF. ⇒ ∠GIE = 90°
Hence, ∠GAB = ∠GIE = 90° [∵ corresponding angles]
Construct an angle of 45° at the initial point of a given ray and justify the construction.
Steps of construction (i) Draw a ray AB at the point A. (ii) Taking A as centre and a convenient radius, draw an arc which intersect AB at C. (iii) Taking C as centre and with the same radius, draw an arc which intersect the previous arc at D. (iv) Similarly, taking D as centre, draw arcs with the sRead more
Steps of construction
See less(i) Draw a ray AB at the point A.
(ii) Taking A as centre and a convenient radius, draw an arc which intersect AB at C.
(iii) Taking C as centre and with the same radius, draw an arc which intersect the previous arc at D.
(iv) Similarly, taking D as centre, draw arcs with the same radius, draw an arc which intersects at E.
(v) taking D and E as centre, draw arcs with equal radius (more than half of DE), Which intersect at F.
(vi) From the point A, draw a ray AF, which intersects arc DE at G.
(vii) Taking C and G as centre. Draw arcs with equal radius (more than half of CG), which intersect at H.
(viii) From the point A, draw a ray of 45°.
Justification: Join GH and CH.
In ΔAGH and ΔACH,
GH = CH [∵ Arcs of equal radii]
AG = AC [∵ Radii of same circle]
AH = AH [∵ Common]
So, ΔAGH ≅ ΔACH [∵ SSS Congruency rule]
∠GAH = ∠CAH [∵ CPCT]
Hence, ∠GAH = ∠CAH = 45°
A triangle and a parallelogram have the same base and the same area. If the sides of the triangle are 26 cm, 28 cm and 30 cm, and the parallelogram stands on the base 28 cm, find the height of the parallelogram.
Here, the sides of triangle ABC are a = 28 cm, b = 26 and c = 30 cm. So, the semi-perimeter of triangle S = a+b+c/2 = 28+26+30+ = 84/2 = 42 cm Therefore, using Heron's formula, area of triangle = √s(s - a)(s -b)(s -c) = √42(42 -28)(42 -26)(42 -30) = √42(14)(16)(12) = √112896 = 366 cm² We know that tRead more
Here, the sides of triangle ABC are a = 28 cm, b = 26 and c = 30 cm.
See lessSo, the semi-perimeter of triangle S = a+b+c/2 = 28+26+30+ = 84/2 = 42 cm
Therefore, using Heron’s formula, area of triangle = √s(s – a)(s -b)(s -c)
= √42(42 -28)(42 -26)(42 -30) = √42(14)(16)(12) = √112896
= 366 cm²
We know that the area of a parallelogram = base × corresponting hieght
According to question:
Area of parallelogram = Area of triangle
⇒ base × corresponding height = 336
⇒ 28 × corresponding height = 336
⇒ corresponding height = 336/28 = 12 cm.
A rhombus shaped field has green grass for 18 cows to graze. If each side of the rhombus is 30 m and its longer diagonal is 48 m, how much area of grass field will each cow be getting?
Join the diagonal AC of quadrilateral ABCD. Here, the sides of triangle ABC are a = 30 m, b = 30 m and c = 48 m. So, the semi-perimeter of triangle S = a + b + c/2 = 30 + 30 + 48/2 = 108/2 = 54 m. Therefore, area of triangle = √s(s - a)(s -b)(s -c) = √54(54 -30)(54 -30)(54 -48) = √54(24)(24)(6) = √1Read more
Join the diagonal AC of quadrilateral ABCD.
See lessHere, the sides of triangle ABC are a = 30 m, b = 30 m and c = 48 m.
So, the semi-perimeter of triangle
S = a + b + c/2 = 30 + 30 + 48/2 = 108/2 = 54 m.
Therefore, area of triangle = √s(s – a)(s -b)(s -c)
= √54(54 -30)(54 -30)(54 -48) = √54(24)(24)(6) = √186624 = 432 m²
Hence, area of quadrilateral = 2 × 432 = 864 m²
Therefore, the area grazed by each cow = Total area/number of cows = 864/18 = 48 m²