Advertisements
Advertisements
Question
PQ is perpendicular to BA and BD is perpendicular to AP.PQ and BD intersect at R. Prove that ΔABD ∼ ΔAPQ and `"AB"/"AP" = "BD"/"PQ"`.
Solution
In ΔABD and ΔAPQ,
∠BDA ∠PQA = 90°
∠A = ∠A
Therefore, ΔABD ∼ ΔAPQ ...(AA axiom)
And hence, `"AB"/"AP" = "BD"/"PQ"`.
APPEARS IN
RELATED QUESTIONS
The diagonals of a quadrilateral ABCD intersect each other at the point O such that `("AO")/("BO") = ("CO")/("DO")`. Show that ABCD is a trapezium.
On a map, drawn to a scale of 1 : 250000, a triangular plot PQR of land has the following measurements :
PQ = 3cm, QR = 4 cm and angles PQR = 90°
(i) the actual lengths of QR and PR in kilometer.
(ii) the actual area of the plot in sq . km.
ABCD is parallelogram and E is a point on BC. If the diagonal BD intersects AE at F, prove that AF × FB = EF × FD.
The corresponding altitudes of two similar triangles are 6cm and 9cm respectively. Find the ratio of their areas.
In the given figure, PQ || AB; CQ = 4.8 cm QB = 3.6 cm and AB = 6.3 cm. Find :
- `(CP)/(PA)`
- PQ
- If AP = x, then the value of AC in terms of x.
In figure, PQ is parallel to BC, AP : AB = 2 : 7. If QC = 0 and BC = 21,
Find
(i) AQ
(ii) PQ
The scale of a map is 1 : 50000. The area of a city is 40 sq km which is to be represented on the map. Find: The length of a scale in km represented by 1cm on the map.
On a map drawn to a scale of 1:25000, a triangular plot of land is right angled and the sides forming the right angle measure 225cm and 64cm.Find: The area of the plot in sq. km.
If ΔABC ~ ΔLMN and ∠B = 40°, then ∠M = ? Give reason.
In fig. BP ⊥ AC, CQ ⊥ AB, A−P−C, and A−Q−B then show that ΔAPB and ΔAQC are similar.
In ΔAPB and ΔAQC
∠APB = [ ]° ......(i)
∠AQC = [ ]° ......(ii)
∠APB ≅ ∠AQC .....[From (i) and (ii)]
∠PAB ≅ ∠QAC .....[______]
ΔAPB ~ ΔAQC .....[______]
