Ex: In the figure, ABC is an equilateral triangle.
(a) Prove that, summation of four angles of a
quadrilaterals equal to four right angles.
Prove that, ∠BDF+∠DFE+∠FEB+∠EBD= 4 right angles.
(b) Prove that, DF∥BC and DF = ½ BC.
Particular enunciation: Given that, in the figure, ABC is an equilateral triangle. D,E and F are the midpoints of AB, BC and AC respectively. We join D, F; E, F and D,E.
We have to prove that ∠BDF+∠DFE+∠FEB+∠EBD = 4 right angle.
Proof: In ∆DBE
∠DBE+∠BED+BDE = 2 right angle. ————— (1) [∵sum of three angles of a triangle is equal to 2 right angle]
Again, In ∆DEF
∠EDF+∠DEF+∠DFE= 2 right angle ———————- (2)
Adding (1) and (2) we get,
Or, ∠DBE+(∠BED+∠DEF)+ ∠EFD+(∠BDE+∠EDF)=4 right angle
Or, ∠DBE+∠BEF+∠EFD+∠BDF=4 right angle.
Therefore ∠BDF+∠DFE+∠FEB+∠EBD=4 right angle (Proved)
Ex: Prove that, DF∥BC and DF= ½ BC
Particular enunciation: given that in the figure, ABC is an equilateral triangle. D, E and F are the mid points of AB, BC and AC respectively. We join D and F.
We have to prove that DE∥BC and DF = ½ BC.
Construction: we produce DF upto O such that DF=FO. We join O and C.
Proof: In ∆ADF and ∆COF
AF=CF [F is midpoint of AC]
DF=OF [by construction]
And included ∠AFD=included ∠CFO [vertically opposite angles are equal to each other]
So ∆ADF≌∆COF [SSS theorem]
Or, BD=CO [[D is an midpoint AB]
But these are alternate angles
The opposite sides BD and CO are equal and parallel.
∴⎕BCOD is a parallelogram.
So DO ∥BC [the opposite of a parallelogram are equal and parallel]
Or, DF ∥BC
Or, DF+DF=BC [DF=FO]
Or, 2 DF=BC
∴DF= ½ BC (Proved)
Ex: In the figure, AB=CD and AB∥CD.
(a)Name the two triangles on base AB.
(b)Prove that AD and BC are equal and
Parallel to each other
(c)Show that, AO=OC and OB=OD.
From the figure, the two triangles on base AB are(i) ∆ABC and (ii)∆ABD.
Particular enunciation : Given that AC=CD. We have to prove that AD and BC are equal and parallel to each other.
Construction: we join A and C.
Proof: AB∥CD and AC is their Transversal.
∴∠BAC=alternative ∠DCA ————- (i) [Alternative angles are equal to each other]
In ∆ABC and ∆ADC
AC=AC [common side]
And included ∠BAC=included ∠DCA [from (i)]
So, ∆ABC≌∆ ADC [SSS theorem]
Now, AC is the transversal of the sides AD and BC and∠ BCA=alternate∠ DAC
We know, if alternate angles are equal then two sides are parallel.
(c) Particular enunciation: given that AB=CD and AB∥CD. We join A,C and B,D.AC and BD intersects each other at O.
We have to prove that OA=OC and OB=OD.
And AD∥BC [from (b)]
∴ABCD is a parallelogram.
In the parallelogram ABCD, the diagonals AC and BD intersect each other at O.
Since, the diagonals of a parallelogram bisect each other
∴AO=OC and OB = OD (Proved)
Ex: Length, breadth and height of a rectangular solid are 10cm, 8cm and 5cm respectively. Find the area of the whole face of the solid.
Length of a rectangular solid a, =10cm
Breadth of a rectangular solid is b,=8cm
Height of a rectangular solid is c,=5cm
Area of the whole face of solid =2 (ab + bc + ca)sq. unit
Ex: If the edge of a cubic box is 6.5 cm, find the area of the entire faces of the box.
The edge of a cubic box, a=6.5 cm
Area of the entire faces of the box=6a2 sq. unit
=253.5 sq.cm (Ans.)