Advertisements
Advertisements
Question
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.
Advertisements
Solution
Given that
AB = 7cm, BC = 8cm, AC = 9cm .
In ΔABC
∴ F and E are the midpoint of AB and AC
∴EF = `1/2` BC [Mid-points theorem]
Similarly
DF = `1/2` AC, DE = `1/2` AB
Perimeter of ΔDEF = DE + EF + DF
= `1/2` AB + `1/2` BC `1/2`AC
= `1/2`× 7 + `1/2` × 8 +` 1 /2`× 9
= 3.5 + 4 + 4.5 = 12cm
∴ Perimeter of ΔDEF = 12cm
APPEARS IN
RELATED QUESTIONS
ABCD is a rectangle and P, Q, R and S are mid-points of the sides AB, BC, CD and DA respectively. Show that the quadrilateral PQRS is a rhombus.
In the Figure, `square`ABCD is a trapezium. AB || DC. Points P and Q are midpoints of seg AD and seg BC respectively. Then prove that, PQ || AB and PQ = `1/2 ("AB" + "DC")`.
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.
D and F are midpoints of sides AB and AC of a triangle ABC. A line through F and parallel to AB meets BC at point E.
- Prove that BDFE is a parallelogram
- Find AB, if EF = 4.8 cm.
In Δ ABC, AD is the median and DE is parallel to BA, where E is a point in AC. Prove that BE is also a median.
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.
In ΔABC, D is the mid-point of AB and E is the mid-point of BC.
Calculate:
(i) DE, if AC = 8.6 cm
(ii) ∠DEB, if ∠ACB = 72°
In the given figure, ABCD is a trapezium. P and Q are the midpoints of non-parallel side AD and BC respectively. Find: PQ, if AB = 12 cm and DC = 10 cm.
ABCD is a parallelogram.E is the mid-point of CD and P is a point on AC such that PC = `(1)/(4)"AC"`. EP produced meets BC at F. Prove that: 2EF = BD.
ABCD is a kite in which BC = CD, AB = AD. E, F and G are the mid-points of CD, BC and AB respectively. Prove that: The line drawn through G and parallel to FE and bisects DA.
