📐
Geometry Construction
Math MMN | Class 10 SEE Prep
Share Notes:
QUESTION 1: Construct Quad ABCD & Equal Area Triangle ADE
(IMPORTANT FOR SEE)
Given: \( AB = BC = 5\text{ cm} \), \( CD = AD = 6\text{ cm} \), Diagonal \( AC = 8\text{ cm} \)
Target: \( \text{Area}(\Delta ADE) = \text{Area}(\text{Quad } ABCD) \) using Parallel \( CE \parallel BD \)
Target: \( \text{Area}(\Delta ADE) = \text{Area}(\text{Quad } ABCD) \) using Parallel \( CE \parallel BD \)
Construction Steps
-
Step 1: Draw Base AB = 5 cm & Extend Ray ABX
Draw line segment \( AB = 5\text{ cm} \) using scale and extend ray \( ABX \). -
Step 2: Locate Point C (BC = 5 cm, AC = 8 cm)
Cut arc of \( 5\text{ cm} \) from \( B \) and \( 8\text{ cm} \) from \( A \) to mark point \( C \). Join \( BC \). -
Step 3: Locate Point D & Quad ABCD
Cut arc of \( 6\text{ cm} \) from \( C \) and \( 6\text{ cm} \) from \( A \) to mark point \( D \). Join \( CD \) and \( AD \). -
Step 4: Join Diagonal BD
Draw diagonal \( BD \) to form reference angle \( \angle DBC \). -
Step 5: Copy Alternate Angle & Draw CE || BD through P
Mark arc \( xy \) at \( \angle DBC \). At \( C \), draw arc to mark \( z \) on \( BC \). Measure arc length \( xy \), then from \( z \) cut arc to get point \( t \). Draw line \( CE \) through \( C \) & \( t \) to locate point \( E \) on line \( ABX \). -
Step 6: Join DE to Get Target Triangle ΔADE
Now let’s join \( DE \) to get triangle \( ADE \) equal in area to quadrilateral \( ABCD \).
Hence, \( \Delta ADE \) is the required triangle such that \( \text{Area}(\Delta ADE) = \text{Area}(\text{Quad } ABCD) \).
QUESTION 2: Construct Quad PQRS & Equal Area Triangle PST
(SEE 2080 Sudurpashchim)
Given: \( PQ = QR = 4.8\text{ cm} \), \( RS = PS = 5.8\text{ cm} \), \( \angle QPS = 60^\circ \)
Target: \( \text{Area}(\Delta PST) = \text{Area}(\text{Quad } PQRS) \) using Parallel \( RT \parallel SQ \)
Target: \( \text{Area}(\Delta PST) = \text{Area}(\text{Quad } PQRS) \) using Parallel \( RT \parallel SQ \)
Construction Steps
-
Step 1: Draw Base Ray PX (PQ = 4.8 cm)
Draw line segment \( PQ = 4.8\text{ cm} \) using scale and extend ray \( PX \). -
Step 2: Construct ∠QPS = 60° & Cut Arc PS = 5.8 cm
Construct angle \( \angle QPS = 60^\circ \) at \( P \) to draw ray \( PY \). Cut arc of \( 5.8\text{ cm} \) on \( PY \) from \( P \) to locate point \( S \). -
Step 3: Locate Point R & Complete Quad PQRS
Draw arc \( QR \) (\( 4.8\text{ cm} \)) from \( Q \) and arc \( SR \) (\( 5.8\text{ cm} \)) from \( S \). Mark \( R \), join \( QR \) & \( RS \) using scale. -
Step 4: Join Diagonal SQ
Draw dashed line \( SQ \) from \( Q \) to \( S \) using scale to form \( \angle SQR \). -
Step 5: Alternate Angle ∠QRT = ∠SQR & Line RT
Measure arc at \( Q \). Place compass at \( w \) on line \( QR \) at \( R \), cut arc \( M \), and draw line \( RM \) to intersect \( PX \) at \( T \). -
Step 6: Target Triangle ΔPST Completed
Join \( ST \) using scale. Shaded area \( \Delta PST \) is the required triangle equal in area to quad \( PQRS \).
Hence, \( \Delta PST \) is the required triangle such that \( \text{Area}(\Delta PST) = \text{Area}(\text{Quad } PQRS) \).
QUESTION 3: Construct Quad FEWA & Equal Area Triangle FAT
(SEE Model Question)
Given: \( FE = 5\text{ cm} \), \( \angle FEW = 90^\circ \), \( \angle WFE = 45^\circ \), \( FA = WA = 6\text{ cm} \)
Target: \( \text{Area}(\Delta FAT) = \text{Area}(\text{Quad } FEWA) \) using Parallel \( WT \parallel EA \)
Target: \( \text{Area}(\Delta FAT) = \text{Area}(\text{Quad } FEWA) \) using Parallel \( WT \parallel EA \)
Construction Steps
-
Step 1: Draw Base Ray FEX (FE = 5 cm)
Draw line segment \( FE = 5\text{ cm} \) using scale and extend ray \( FEX \). -
Step 2: Construct 90° at E (Points r1, s1, t1) & Ray EY
Construct \( 90^\circ \) at \( E \). Mark \( 60^\circ \) at \( r1 \), \( 120^\circ \) at \( s1 \), and top intersection \( t1 \) to draw perpendicular ray \( EY \). -
Step 3: Construct 90° at F (Points r2, s2, t2) & Bisect for 45°
Construct \( 90^\circ \) at \( F \) with \( r2, s2, t2 \). Bisect \( 0^\circ \) and \( 90^\circ \) to draw \( 45^\circ \) ray intersecting \( EY \) at \( W \). -
Step 4: Locate Point A (FA = WA = 6 cm)
Cut arcs of \( 6\text{ cm} \) from \( F \) and \( W \) to locate Point \( A \). Join \( FA \) and \( WA \). -
Step 5: Copy Alternate Angle ∠AEW at W (Red Dashed W…
Join \( EA \). Draw arc at \( E \) (\( p, q \)) & base arc at \( W \) (\( p’ \)). Cut red arc from \( p’ \) using length \( pq \) to form \( q’ \), then draw line \( WT \) through \( W \) and \( q’ \). -
Step 6: Target Triangle ΔFAT Completed
Join \( AT \). Shaded area \( \Delta FAT \) is the required triangle equal in area to quadrilateral \( FEWA \).
Hence, \( \Delta FAT \) is the required triangle such that \( \text{Area}(\Delta FAT) = \text{Area}(\text{Quad } FEWA) \).