Advertisements
Advertisements
Question
Prove that the figure obtained by joining the mid-points of the adjacent sides of a rectangle is a rhombus.
Advertisements
Solution
Given: Let ABCD be a rectangle where P, Q, R, S are the midpoint of AB, BC, CD, DA.
To Prove: PQRS is a rhombus
Construction: Draw two diagonal BD and AC as shown in figure. Where BD = AC
(Since diagonal of the rectangle are equal)
Proof:
From ΔABD and ΔBCD
PS = `1/2` BD = QR and PS || BD || QR
2PS = 2QR = BD and PS || QR ...(1)
Similarly, 2PQ = 2SR = AC and PQ || SR ...(2)
From (1) and (2) we get
PQ = QR = RS = PS
Therefore, PQRS is a rhombus.
Hence, proved.
APPEARS IN
RELATED QUESTIONS
In below fig. ABCD is a parallelogram and E is the mid-point of side B If DE and AB when produced meet at F, prove that AF = 2AB.
ABCD is a square E, F, G and H are points on AB, BC, CD and DA respectively, such that AE = BF = CG = DH. Prove that EFGH is a square.
In a triangle ∠ABC, ∠A = 50°, ∠B = 60° and ∠C = 70°. Find the measures of the angles of
the triangle formed by joining the mid-points of the sides of this triangle.
Fill in the blank to make the following statement correct
The triangle formed by joining the mid-points of the sides of an isosceles triangle is
Fill in the blank to make the following statement correct:
The triangle formed by joining the mid-points of the sides of a right triangle is
In the given figure, seg PD is a median of ΔPQR. Point T is the mid point of seg PD. Produced QT intersects PR at M. Show that `"PM"/"PR" = 1/3`.
[Hint: DN || QM]
In parallelogram ABCD, E and F are mid-points of the sides AB and CD respectively. The line segments AF and BF meet the line segments ED and EC at points G and H respectively.
Prove that:
(i) Triangles HEB and FHC are congruent;
(ii) GEHF is a parallelogram.
D, E and F are the mid-points of the sides AB, BC and CA of an isosceles ΔABC in which AB = BC. Prove that ΔDEF is also isosceles.
In ΔABC, the medians BE and CD are produced to the points P and Q respectively such that BE = EP and CD = DQ. Prove that: Q A and P are collinear.
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.
