Advertisements
Advertisements
Question
In Fig. below, triangle ABC is right-angled at B. Given that AB = 9 cm, AC = 15 cm and D,
E are the mid-points of the sides AB and AC respectively, calculate
(i) The length of BC (ii) The area of ΔADE.
Advertisements
Solution
In right ΔABC, ∠B = 90°
By using Pythagoras theorem
`AC^2 = AB^2+ BC^2`
⇒ `15^2 = 9^2 +BC^2`
⇒ BC =`sqrt(15^2 - 9^2)`
⇒ BC =`sqrt(225-81)`
⇒ BC =`sqrt144`
= 12cm
In ΔABC
D and E are midpoints of AB and AC
∴ DE || BC, DE = `1/2` BC [By midpoint theorem]
AD = OB = `(AB)/ 2= 9/2` = 4 . 5cm [ ∵ D is the midpoint of AB]
DE = `(BC)/2 = 12/2` = 6cm
Area of ΔADE = `1/2 xxAD xx DE `
= `1/2× 4 .5 × 6 = 13.5cm^2`
APPEARS IN
RELATED QUESTIONS
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.
A parallelogram ABCD has P the mid-point of Dc and Q a point of Ac such that
CQ = `[1]/[4]`AC. PQ produced meets BC at R.
Prove that
(i)R is the midpoint of BC
(ii) PR = `[1]/[2]` DB
In triangle ABC, P is the mid-point of side BC. A line through P and parallel to CA meets AB at point Q, and a line through Q and parallel to BC meets median AP at point R.
Prove that : (i) AP = 2AR
(ii) BC = 4QR
In triangle ABC, angle B is obtuse. D and E are mid-points of sides AB and BC respectively and F is a point on side AC such that EF is parallel to AB. Show that BEFD is a parallelogram.
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.
AD is a median of side BC of ABC. E is the midpoint of AD. BE is joined and produced to meet AC at F. Prove that AF: AC = 1 : 3.
In a parallelogram ABCD, E and F are the midpoints of the sides AB and CD respectively. The line segments AF and BF meet the line segments DE and CE at points G and H respectively Prove that: ΔHEB ≅ ΔHFC
In ΔABC, D and E are the midpoints of the sides AB and AC respectively. F is any point on the side BC. If DE intersects AF at P show that DP = PE.
D and E are the mid-points of the sides AB and AC of ∆ABC and O is any point on side BC. O is joined to A. If P and Q are the mid-points of OB and OC respectively, then DEQP is ______.
