Advertisements
Advertisements
प्रश्न
The following figure shows a trapezium ABCD in which AB // DC. P is the mid-point of AD and PR // AB. Prove that:
PR = `[1]/[2]` ( AB + CD)
उत्तर
Here from the triangle,
ABD P is the midpoint of AD and PR || AB,
therefore Q is the midpoint of BD
Similarly, R is the midpoint of BC as PR || CD || AB
From triangle ABD,
PQ = `1/2` AB ...(1) ...[by Mid-point theorem]
From triangle BCD,
QR = `1/2` CD ...(2) ...[by Mid-point theorem]
Now (1) + (2)
PQ + QR = `1/2 "AB" + 1/2 "CD"`
PR = `1/2`(AB + CD)
Hence proved.
APPEARS IN
संबंधित प्रश्न
Show that the line segments joining the mid-points of the opposite sides of a quadrilateral bisect each other.
In a ∆ABC, D, E and F are, respectively, the mid-points of BC, CA and AB. If the lengths of side AB, BC and CA are 7 cm, 8 cm and 9 cm, respectively, find the perimeter of ∆DEF.
ABC is a triangle and through A, B, C lines are drawn parallel to BC, CA and AB respectively
intersecting at P, Q and R. Prove that the perimeter of ΔPQR is double the perimeter of
ΔABC
The figure, given below, shows a trapezium ABCD. M and N are the mid-point of the non-parallel sides AD and BC respectively. Find:
- MN, if AB = 11 cm and DC = 8 cm.
- AB, if DC = 20 cm and MN = 27 cm.
- DC, if MN = 15 cm and AB = 23 cm.
ABCD is a quadrilateral in which AD = BC. E, F, G and H are the mid-points of AB, BD, CD and Ac respectively. Prove that EFGH is a rhombus.
In parallelogram ABCD, E is the mid-point of AB and AP is parallel to EC which meets DC at point O and BC produced at P.
Prove that:
(i) BP = 2AD
(ii) O is the mid-point of AP.
In the given figure, AD and CE are medians and DF // CE.
Prove that: FB = `1/4` AB.
In the given figure, PS = 3RS. M is the midpoint of QR. If TR || MN || QP, then prove that:
RT = `(1)/(3)"PQ"`
D, E and F are the mid-points of the sides BC, CA and AB, respectively of an equilateral triangle ABC. Show that ∆DEF is also an equilateral triangle.
P and Q are the mid-points of the opposite sides AB and CD of a parallelogram ABCD. AQ intersects DP at S and BQ intersects CP at R. Show that PRQS is a parallelogram.
