Advertisements
Advertisements
प्रश्न
If figure OPRQ is a square and ∠MLN = 90°. Prove that ∆QMO ~ ∆RPN
उत्तर
In ∆QMO and ∆RPN
∠MQO = ∠NRP = 90°
∠RPN = ∠QOM ...(OP || MN)
∴ ∆QMO ~ ∆RPN ...(By AA similarity)
APPEARS IN
संबंधित प्रश्न
The given figure shows a triangle PQR in which XY is parallel to QR. If PX : XQ = 1 : 3 and QR = 9 cm, find the length of XY.
Further, if the area of ΔPXY = x cm2; find, in terms of x the area of :
- triangle PQR.
- trapezium XQRY.
The areas of two similar triangles are `64cm^2` and `100cm^2` respectively. If a median of the smaller triangle is 5.6cm, find the corresponding median of the other.
In the given figure, X is any point in the interior of triangle. Point X is joined to vertices of triangle. Seg PQ || seg DE, seg QR || seg EF. Fill in the blanks to prove that, seg PR || seg DF.
Proof : In ΔXDE, PQ || DE ...`square`
∴ `"XP"/square = square/"QE"` ...(I) (Basic proportionality theorem)
In ΔXEF, QR || EF ...`square`
∴ `square/square = square/square ..."(II)" square`
∴ `square/square = square/square` ...from (I) and (II)
∴ seg PR || seg DF ...(converse of basic proportionality theorem)
In Δ PQR, MN is drawn parallel to QR. If PM = x, MQ = (x-2), PN = (x+2) and NR = (x-1), find the value of x.
In the given figure, PQ || AB; CQ = 4.8 cm QB = 3.6 cm and AB = 6.3 cm. Find : If AP = x, then the value of AC in terms of x.
In fig.DE || BC ,AD = 1 cm and BD = 2 cm. what is the ratio of the ar(ΔABC) to the ar (ΔADE)?
In the figure, AB || RQ and BC || SQ, prove that `"PC"/"PS" = "PA"/"PR"`.
Harmeet is 6 feet tall and casts a shadow of 3 feet long. What is the height of a nearby pole if it casts a shadow of 12 feet long at the same time?
A girl looks the reflection of the top of the lamp post on the mirror which is 6.6 m away from the foot of the lamppost. The girl whose height is 1.25 m is standing 2.5 m away from the mirror. Assuming the mirror is placed on the ground facing the sky and the girl, mirror and the lamppost are in the same line, find the height of the lamp post.
Prove that if a line is drawn parallel to one side of a triangle intersecting the other two sides in distinct points, then the other two sides are divided in the same ratio.
Using the above theorem prove that a line through the point of intersection of the diagonals and parallel to the base of the trapezium divides the non-parallel sides in the same ratio.
