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Syllabus

Q.74. In the parallelogram ABCD, M is mid-point of AC and X, Y are points on AB and DC respectively such that AX = CY.

Prove that :

(i) triangle AXM is congruent to triangle CYM,

(ii) KMY is a straight line.

Prove that

I) angle PQR = 90 DEGREE

II) Line through P and parallel to QR bisects side AD

^{0.5}: 1 ( 3^{0.5}= square root of 3)Q. In triangle PQR, MN||QR and $\frac{PM}{MQ}=\frac{2}{3}$

(i) Find $\frac{MN}{QR}$

(ii) Prove that triangle OMN and ORQ are similar

(iiI) Find, area of triangle OMN : area of triangle ORQ

Please do not provide any link.

1) AC is greater than AD

2) AE is greater than AC

3) AE is greater than AD

If AO = 2CO and BO = 2DO; show that:

(i) ∆AOB is similar to ∆COD.

(ii) OA × OD = OB × OC.

(i) the actual length of the diagonal distance AC of the plot in km.

(ii) the actual area of the plot sq km.

1. the actual length of the diagonal distance AC of the plot in km.

2. the actual area of the plot in sq km.

(i) the length of PQ, if BC=7.5cm.

(ii)the area of triangle APQ: area of triangle ABC

(iii)the area of triangleAPQ:area of PBCQ

(b) In $\u2206$PQR, MN is parallel to QR and

$\frac{PM}{MQ}=\frac{2}{3}$

(i) Find

$\frac{MN}{QR}$

(ii) Prove that $\u2206$OMN and

$\u2206$ORQ are similar.

(iii) Find , Area of $\u2206$OMN : Area of $\u2206$ORQ

1) M , N and A are collinear.

2) A is the mid point of MN.